After studying the chapter Factorization[class 9], you learn to find the factors of simple monomial expressions, find the factors of binomial expressions, find the factors of some cubic expressions.

## 3.2.1 Introduction to Factorization[class 9]:

### Different methods of Factorization[class 9]:

**Method 1: Factorization[class 9] taking common factors:**

When each term of an expression has a common factor we divide each term by this and take it out as shown below:

**Example: Factorise 9x ^{2} + 12xy**

Solution:

We write 9x^{2} + 12xy = (3x)(3x) + (3x)(4y)

Here 3x is the common to both the terms. Hence,

(3x)(3x) + (3x)(4y) = 3x(3x + 4y)

Thus, 3, x, (3x + 4y) are the factors of 9x^{2} + 12xy

#### Method 2: Factorization[class 9] by grouping:

Sometimes, in a given expression it is not possible to take out a common factor directly. However the terms of the given expression are grouped in such a manner that we have a common factor

**Example: Factorise ab + bc + ax + bx**

Solution:

We have ab + bc + ax + cx = (ab + bc) +(ax +cx) = b(a + c) + x(a+ c) = (a + c)(b + x)

## Factorization[class 9] Exercise 3.2.1

**Factorise:**

**(i) 5x ^{2}**

**– 20xy**

Solution:

= 5x(x – 4y)

**(ii) 2p(x + y) – 3q(x + y) **

Solution:

= (x + y) (2p – 3q)

**(iii) 4(a + b) ^{2}**

**– 6(a + b)**

Solution:

= 2 (a + b) [2(a + b) – 3]

= 2 (a + b) (2a + 2b – 3)

**(iv) 18x ^{2}y – 24xyz **

Solution:

= 6xy (3x – 4z)

**(v) 7xa – 70 xb **

Solution:

= 7x (a – 10b)

**(vi) 13m ^{2}**

**+ 156n**

^{2}Solution:

=13(m^{2} + 12n^{2})

**(vii) 3x ^{2}**

**+ 6x +6**

Solution:

= 3 (x^{2} + 2x + 2)

**(viii) 4abx ^{2}**

**+ 8abx + 120by**

Solution:

= 4ab (x^{2} + 2x +3y)

**2. Factorise **

**(i) x ^{2}**

**+ xy + xz + yz**

Solution:

= x (x + y) + z (x+ y)

=(x + y) (x + z)

**(ii) a ^{2}**

**– ab + ac – bx**

Solution:

= a (a – b) + c (a – b)

= (a – b) (a – c)

(**iii) 4x + bx + 4b + b ^{2}**

Solution:

= x (4 + b) + b (4 + b)

= (4 + b) (x + b)

**(iv) 6ax – 6a + 5a – 5 **

Solution:

= 6a (x – 1) + 5 (x – 1)

= (3x + 5) (x^{2} + 1)

**(v) a ^{3}**

**+ a**

^{2}b + ab + b^{2}Solution:

= a^{2} (a +b) + b (a +b)

= (a + b) (a^{2} + b)

**(vi) 3x ^{3}**

**+ 5x**

^{2 }+ 3x +5Solution:

= x^{2} (3x +5) + 1 (3x + 5)

= (3x + 5) (x^{2} + 1)

**(vii) y ^{4}**

**– 2y**

^{3}**+ y – 2**

Solution:

= y^{3}(y – 2) + 1 (y – 2)

= (y – 2) (y^{3} – 1)

= (y – 2) (y^{3} – 1^{3})

= (y – 2) (y + 1) (y^{2} – y +1)

**(viii) t ^{4}**

**+ t**

^{4}**– 2t – 2**

Solution:

= t^{3} (t + 1) – 2(t +1)

= (t – 1) (t^{3} – 2)

## 3.2.2 Factorization of the difference of two square – Factorization[class 9]:

We know that (a + b)(a – b) = a^{2} – b^{2}. We can use this identity to factorise many expressions involving the difference of the two squares.

**Example 8: Factorise 25x ^{2} – 64y^{2}**

Solution:

We write 25x^{2} – 64y^{2}

25x^{2} – 64y^{2} = (5x)^{2} – (8y)^{2}

This is in the form a^{2} – b^{2} which factorises to (a – b)(a + b). Thus,

25x^{2} – 64y^{2} = (5x)^{2} – (8y)^{2} = (5x – 8y)(5x + 8y)

## Factorization[class 9] – Exercise 3.2.1

**Factorize the following:**

**(i) 9x**^{2}**– 16y**^{2}

Solution:

= (3x)^{2} – (4y)^{2}

Using a^{2} – b^{2} = (a + b) (a – b)

= (3x + 4y) (3x – 4y)

**(ii) 5x**^{2}**– 7y**^{2}**[ Hint : ( ****𝟓 ****)**^{2}**= 5 ] **

Solution:

( 5 x)^{2} – ( 7 y)^{2}

Using a^{2} – b^{2} = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)

**(iii) 100 – 9x****2 **

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a^{2} – b^{2} = (a + b) (a – b)]

**(iv) (x + 4y)**^{2}**– 4z**^{2}

Solution:

= (x + 4y)^{2} – (2z)^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)

**(v) x – 64xy**^{4}

Solution:

= x (1 – 64y^{4})

= x (1 + (8y^{2})^{2}) [∵a^{2} – b^{2} = (a + b) (a – b)]

= x (1 + 8y^{2}) (1 – 8y^{2})

**(vi) a**^{2}**+ 2ab + b**^{2}**– 9c**^{2}

Solution:

= (a + b)^{2} – (3c)^{2} [∵(a + b)^{2} = a^{2} + 2ab + b^{2} ;a^{2} – b^{2} = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)

**(vii) 4x**^{2}**– 9y**^{2}**– 2x – 3y **

Solution:

= (2x)^{2} – (3y)^{2} – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a^{2} – b^{2 }= (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)

**(viii) a**^{2}**+ b**^{2}**– a + b **

Solution:

= a^{2} + b^{2 }– (a – b) [∵a^{2} – b^{2} = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)

**(ix) x**^{4}**– 625 **

Solution:

= (x^{2})^{2} – 25^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 25) (x^{2} – 25)

= (x^{2} + 25) (x^{2} – 5^{2})

= (x^{2} + 25) (x + 5) (x – 5)

**(x) 36 (5x + y)**^{2}**– 25 (4x – y)**^{2 }

Solution:

= [6 (5x + y)]^{2} – [5 (4x – y)]^{2} [∵a^{2 }– b^{2} = (a + b) (a – b)]

= (30x + 6y + 20x – 5y) (30x + 6y – 20x + 5y)

= (50x + y) (10x + 11y)

**(xi) (a + **^{𝟏}**/ _{𝐚} **

**)**

^{2}**– 4 (x –**

^{𝟏}**/**

_{𝐱}**)**

^{2}Solution:

= (a + ^{𝟏}/_{𝐚} )^{2} – [2 (x – ^{𝟏}/_{𝐱} )]^{2}

= [a+ ^{𝟏}/_{𝐚} + 2( x− ^{𝟏}/_{𝐱})][ a+ ^{𝟏}/_{𝐚} – 2(x− ^{𝟏}/_{𝐱} )]

**(xii) 12m**^{2}**– 75n**^{2}

Solution:

= ( √12 m^{2})^{2} – (√75 n^{2})^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (2 √3m^{2})^{2} – (5√3 n^{2})^{2 }[ 12 =2√3 and 75 = 5√3

= (2√3m^{2} + 5√3 n^{2}) (2√3 m^{2} – 5√3 n^{2})

= √3 (2m^{2} + 5n^{2}) (2m – 5n) (2m + 5n)

**Factorize by adding and subtracting appropriate quantity.**

**(i) x**^{2}**+ 6x + 3 **

Solution:

= x^{2} + ^{2}. x. 3 + 3^{2} – 3^{2} + 3

= (x + 3)^{2} – 9 + 3

= (x + 3)^{2} – 6 [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 3)^{2} – 6^{2}

= (x + 3 + 6 ) (x + 3 – 6 )

**(ii) x**^{2}**+ 10x + 8 **

Solution:

= x^{2} + 2. x. 5 + 5^{2} – 5^{2} + 8

= (x + 5)^{2} – 25 + 8

= (x + 5)^{2} – 17 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 5)^{2} – 17^{2}

= (x + 5 + 17 ) (x + 5 – 17 )

**(iii) x**^{2}**+ 10x + 20 **

Solution:

= x^{2} + 2. x. 5 + 5^{2} – 5^{2} + 20 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 5)^{2} – 25 + 20

= (x + 5)^{2} – 5

= (x + 5)^{2} – 5^{2} [∵a^{2} – b^{2} = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )

**(iv) x**^{2}**+ 2x – 1 **

Solution:

= x^{2} + 2. x. 1 + 1^{2} – 1^{2} + 1 [∵a^{2} + 2ab + b^{2} = (a + b)^{2}]

= (x + 1)^{2} – 1 + 1

= (x + 1)^{2} – 2

= (x + 1)^{2} – 2^{2} [∵a^{2} – b^{2 }= (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )

**prove that (1 +**^{𝐱}**/**_{𝐲}**) (1 –**^{𝐱}**/**_{𝐲}**) (1 +****𝐱**^{𝟐}**/****𝐲**^{𝟐}**) (1 +****𝐱**^{𝟒}**/****𝐲**^{𝟒}**) +****𝐱**^{𝟖}**/𝐲**^{𝟖}**= 1**

Solution:

L.H.S = (1 +^{ 𝐱}/_{𝐲} ) (1 – ^{𝐱}/_{𝐲} ) (1 + 𝐱^{𝟐}/𝐲^{𝟐}) (1 + 𝐱^{𝟒}/𝐲^{𝟒} ) + 𝐱^{𝟖}/𝐲^{𝟖}

[∵a^{2} – b^{2} = (a + b) (a – b)]

= (1 – 𝐱^{𝟐}/𝐲^{𝟐} ) (1 + 𝐱^{𝟐}/𝐲^{𝟐} ) (1 + 𝐱^{𝟒}/𝐲^{𝟒}) + 𝐱^{𝟖}/𝐲^{𝟖}

= (1 – 𝐱^{𝟒}/𝐲^{𝟒}) (1 + 𝐱^{𝟒}/𝐲^{𝟒} ) + 𝐱^{𝟖}/𝐲^{𝟖}

= (1)^{2} – (𝐱^{𝟒}/𝐲^{𝟒})^{2} + 𝐱^{𝟖}/𝐲^{𝟖}

= 1 – 𝐱^{𝟖}/𝐲^{𝟖} + 𝐱^{𝟖}/𝐲^{𝟖}

= 1 R.H.S

**If x****=**^{𝐚+𝐛}**/**_{𝐚−𝐛}**and y****=**^{𝐚−𝐛}**/**_{𝐚+𝐛}**find x**^{2}**– y**^{2}**.**

Solution:

x^{2} – y^{2} = (x + y) (x – y)

= (^{a+b}/_{a−b}+^{a−b}/_{a+b})(^{a+b}/_{a−b }– ^{a−b}/_{a+b})

= [[(a+b)^{2}+ (a−b)^{2} ]/(a^{2}− b^{2})][[(a+b)^{2}− (a−b)^{2}]/(a^{2}− b^{2})]

= {(a^{2}+ 2ab + b^{2}+ a^{2}+ b^{2}− 2ab)/[(a^{2}− b^{2})^{2}]} x (a^{2} + 2ab + b^{2} – a^{2} – b2 + 2ab)

= [(2a^{2}+ 2b^{2})4ab]/[(a^{2}− b^{2})^{2}]

= [2(a^{2}+ b^{2})4ab]/(a^{2}− b^{2})^{2}

= [8ab(a^{2}+ b^{2})]/(a^{2}− b^{2})^{2}

**If p = x – y, q = y – z and r = z – x, simplify r**^{2}**– p + 2pq – q**^{2}

Solution:

(z – x)^{2} – (x – y)^{2} + 2 (x – y) (y – z) – (y – z)^{2}

= z^{2} – 2xy + x^{2} – (x^{2} – 2xy + y^{2}) + 2 (xy – y^{2} – xz + yz) – (y^{2} + z^{2} – 2yz)

= z^{2} + x^{2} – 2xz – x^{2} + 2xy – y^{2} + 2xy – 2y^{2} – 2xz + 2yz – y^{2} – z^{2} + 2yz

= 4xy – 4xz + 4yz – 4y^{2}

= 4[xy – xz + yz – y^{2}]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq

**The radius of a circle is 13cm in which a chord of 10cm is drawn. Find the distance of the chord from the centre of the circle.**

**[Hint: use Pythagoras rule and the fact that the perpendicular line drawn from the centre of the circle to any o the chords bisects the chord] **

Solution:

Let AC = 10cm is the chord of the circle with radius 13cm then using Pythagoras theory

OA^{2} = OB^{2} + AB^{2}

13^{2} = x^{2} + 5^{2}

x^{2} = 13^{2} – 5^{2}

= (13 + 5) (13 – 5)

= 18 (8)

x^{2} = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm

**Is it possible to factorize x**^{2}**+ 4x + 20? Give reason**

Solution:

x^{2} + 4x + 20

= x^{2} + 2. x. 2 + 2^{2} – 2^{2} + 20

= x^{2} + 4x + 4 – 4 + 20

= (x + 4)^{2} + 16

= (x + 4)^{2} + 4^{2}

We cannot factorize their further since this is nit of the form a^{2} – b^{2}

## 3.2.3 More on Factorization of Trinomials – Factorization[class 9]:

**Example: Factorize x ^{2} + 4√2x + 8**

Solution:

We write this in the form

x^{2} + 2(x)(2√2) + (2√2)^{2}

This is in the form a^{2} + 2ab + b^{2} , where a = x, b = 2√2

But we know that a^{2} + 2ab + b^{2} = (a + b)^{2}

Thus,

x^{2} + 4√2x + 8 = (x + 2√2)^{2}

**Example: Factorise x ^{2} + 11x + 30**

Solution:

Note that 11 = 5 + 6 and 30 = 5 x 6

This helps us to split the trinomial as,

x^{2} + 11x + 30 = x^{2} + 5x + 6x + 30

= x(x + 5)+6(x + 5)

= (x + 6)(x + 5)

Thus (x+ 5) and (x + 6) are the factors of x^{2} + 11x + 30

**Example: Factorise (x ^{2} – 2x)^{2} – 23(x^{2} – 2x) + 120**

Solution:

We introduce a = x^{2} – 2x . Then the expression is a^{2} – 23a + 120. We observe that – 23 = (-15) + (-8) and 120 = (-15)(-8). Hence we can write a^{2} – 23a + 120 = a^{2} – 15a – 8a + 120 = a(a – 15) – 8(a – 15) = (a – 15)(a – 8)

Hence, (x^{2} – 2x)^{2} – 23(x^{2} – 2x) + 120 = (x^{2} – 2x – 15)( x^{2} – 2x – 8)

Further we see,

(x^{2} – 2x – 15) = x^{2} – 5x + 3x – 15 = x(x – 5)+3(x -5) = (x + 3)(x – 5)

( x^{2} – 2x – 8) = x^{2} – 4x + 2x – 8 = x(x – 4)+2(x – 4) = (x – 4)(x + 2)

We hence obtain

(x^{2} – 2x)^{2} – 23(x^{2} – 2x) + 120 = (x + 3)(x – 5)(x + 2)(x – 4)

## Factorization[class 9] – Exercise 3.2.3

**Factorize**

**(i) x**^{2}**+ 9x + 18 **

Solution:

= x^{2} + 6x + 3x + 18

= x (x + 6) + 3 (x + 6)

= (x + 6) (x + 3)

**(ii) y**^{2}**+ 5y – 24 **

Solution:

= y^{2} + 8y – 3y – 24

= y (y + 8) – 3 (y + 8)

= (y + 8) (y – 3)

**(iii) 7y**^{2}**+ 49y +84 **

Solution:

= 7 (y^{2} + 7y + 12)

= 7 (y^{2} + 4y + 3y + 12)

=7 2018

=7 (y + 4) (y + 3)

**(iv) 40 + 3x – x**^{2}

Solution:

= – [x^{2} – 3x – 40]

= 3 – [x^{2} – 8x + 5x – 40]

= – [x (x – 8) + 5 (x – 8)]

= [(x – 8) (x + 5)]

= (8 – x) (x + 5)

**(v) m**^{2}**+ 17mn – 84n**^{2 }

Solution:

= m^{2} + 21mn – 4mn – 84n^{2}

= m (m + 21n) – 4n (m + 21n)

= (m + 21m) (m – 4n)

**(vi) 117p**^{2 }**+ 2pq – 24q**^{2 }

Solution:

= 117p^{2} + 54pq – 52pq – 24q^{2}

= 9p (13p + 6q) – 3q (13p + 6q)

= (9p – 3q) (13p + 6q)

**(vii) 15x**^{2}**– r – 28 **

Solution:

= 15x^{2} – 21x + 20x – 28

= 3x (5x – 7) + 4 (5x – 7)

= (3x + 4) (5x – 7)

**(viii) 2x**^{2}**– x – 21 **

Solution:

= 2x^{2} – 7x + 6x – 21

= x (2x – 7) + 3 (2x – 7)

= (2x – 7) (x + 3)

**(ix) 8k**^{2}**– 22k – 21 **

Solution:

= 8k^{2} – 28k + 6k – 21

= 4k (2k – 7) + 3(2k – 7)

= (2k – 7) (4k + 3)

**(x) **^{𝟏}**/ _{𝟑} **

**x**

^{2}**– 2x – 9**

Solution:

= (x^{2}+ 6x−27)/_{3}

= ^{1}/_{3} [x^{2} – 6x – 27]

= ^{1}/_{3} [x^{2} – 9x + 3x – 27]

= ^{1}/_{3} [x (x – 9) + 3 (x – 9)]

=^{1}/_{3} (x – 9) (x + 3)

**Factorize**

**(i) √****𝟓 ****x**^{2}**+ 2x – 3√****𝟓 **

Solution:

= √5 x2 + 5x – 3x – 3√5

= √5 x (x + √5 ) – 3 ((x +√ 5 )

= (x + √5 ) (√5 x – 3)

**(ii) √****𝟑 ****a**^{2}**+ 2a – 5√****𝟑 **

Solution:

= √3a^{2} + 5a – 3a – 5√3 – 5√3 – √3

= a (√3 a + 5) – √3 (√3 a + 5)

= (√3 a + 5) (a – √3 )

**(iii) 7√****2 ****y**^{2}**– 10y – 4√****2 **

Solution:

= 7√2 y^{2} – 14y + 4y – 4√2

= 7√2 y (y – √2) + 4 (y – 2)

= (y – √2) (7√2 y + 4)

**(iv) 6√****𝟑 ****z**^{2}**– 47z – 5√****𝟑 **

Solution:

= 6√3 z^{2 }– 45z – 2z – 5√3

= 3√3 z (2z –5√3) – (2z – 5√3)

= (2z – 5√3) (3√3z –1)

**(v) 4√****𝟑****x**^{2}**+ 5x – 2√****𝟑 **

Solution:

= 4√3 x^{2} + 8x – 3x – 2√3

= 4x (√3x + 2) – √3(√3x + 2)

= (√3x + 2) (4x –√3)

**Factorize**

**(i) 2 (x + y)**^{2}**– 9 (x + y) – 5 **

Solution:

Let x + y = p

= 2p^{2} – 9p – 5

= 2p^{2} – 10p + p – 5

= 2p(p – 5) + 1 (p – 5)

= (p – 5) (2p + 1)

= (x + y – 5) [2(x + y) + 1]

= (x + y – 5) (2x + 2y + 1)

**(ii) 2(a – 2b)**^{2}**– 25(a – 2b) + 12 **

Solution:

Put a – 2b = x

= 2x^{2} – 25x + 12

= 2x^{2} – 24x – x + 12

= 2x (x – 12) – 1 (x – 12)

= (x – 12) (2x – 1)

= (a – 2b – 12) [2(a – 2b) – 1]

= (a – 2b – 12) [2a – 4b – 1]

**(iii) 12(z + 1)**^{2}**– 25(z + 1) (x + 2) + 12 (x + 2)**^{2}

Solution:

Let z + 1 = a x + 2 = b

= 12a^{2} – 25ab + 12b^{2}

= 12a^{2} – 16ab – 9ab + 12b^{2}

= 4a (3a – 4b) – 3b (3a – 4b)

= (3a – 4b) (4a – 3b)

= [3 (z + 1) – 4 (x + 2)] [4 (z + 1) – 3 (x + 2)]

= (3z + 3 – 4x – 8) (4z + 4 3x – 2)

= (3z – 4x – 5) (4z – 3x – 2)

**(iv) 9(2x – y) – 4(2x – y) – 13 **

Solution:

9(2x – y) – 4(2x – y) – 13

Put 2x – y = a

= 9a + 9a – 13a – 13

= 9a (a + 1) – 13 (a +1)

= (a + 1) (9a – 13)

= (2x – y + 1) [9 (2x – y) – 13]

= (2x – y + 1) (18x – 9y – 13)

**Factorize**

**(i) x**^{4}**– 3x**^{2}**+ 2 **

Solution:

Put x^{2} = a

a^{2} – 3a + 2

= a^{2} – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 2) (a – 1)

= (x^{2} – 2) (x^{2} – 1)

= (x – 2 ) (x + 2 ) (x – 1) (x + 1) [∵a^{2} – b^{2} = (a + b) (a – b)]

**(ii) 4x**^{4}**+ 7x**^{2}**– 2 **

Solution:

Put x^{2} = a

= 4a^{2} + 7a – 2

= 4a^{2} + 8a – a – 2

= 4a (a + 2) – 1 (a + 2)

= (a + 2) (4a – 1) [∵a^{2 }– b^{2} = (a + b) (a – b)]

= (x^{2} + 2) (4x^{2} – 1)

= (x^{2} + 2) (2x + 1) (2x – 1)

**(iii) 3x**^{3}**– x**^{2}**– 10x **

Solution:

= x (3x^{2} – x – 10)

= x (3x^{2} – 6x + 5x – 10)

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5)

**(iv) 8x**^{3}**+ 2x**^{2}**y – 15xy**^{2}

Solution:

= x (8x^{2} – 2xy – 15y^{2})

= x (8x^{2} – 12xy + 10xy – 15y^{2})

= x [4x (2x – 3y) + 5y (2x – 3y)]

= x (2x – 3y) (4x + 5y)

**(v) x**^{6}**– 7x**^{3}**– 8 **

Solution:

Put x^{3} = a

a^{3} – 7a – 8 [∵a^{3} + b^{3} = (a + b) (a^{2 }– ab + b^{2})]

= a^{3} – 8a + a – 8 [∵a^{3} – b^{3} = (a – b) (a^{2} – ab + b^{2})]

= a (a – 8) + 1 (a – 8)

= (a – 8) (a + 1)

= (x^{3} – 3) (x^{3} + 1)

= (x^{3 }– 3) (x^{3} + 1)

= (x + 1) (x^{2} – x + 1) (x – 2) (x^{2} + 2x + 4)

## 3.2.4 Some Miscellaneous Factorization – Factorization[class 9]:

Consider the expression a^{4} + a^{2}b^{2} + b^{4} .

a^{4} + a^{2}b^{2} + b^{4} = (a^{2} + b^{2} + ab)(a^{2} + b^{2} – ab)

**Example: Factorize 4x ^{4} + 9y^{4} + 6x^{2}y^{2}**

Solution:

a = √2x , b = √3y we obtain,

a^{4} + a^{2}b^{2} + b^{4} = 4x^{4} + 6x^{2}y^{2} + 9y^{4}

We can use the factorization a^{4} + a^{2}b^{2} + b^{4} = (a^{2} + b^{2} + ab)(a^{2} + b^{2} – ab)

4x^{4} + 6x^{2}y^{2} + 9y^{4} = (2x^{2} – √6xy + 3y^{2})(2x^{2} + √6xy + 3y^{2})

## Factorization[class 9] – Exercise 3.2.4

**Resolve into factors**

**(i) x**^{4 }**+ y**^{4}**– 7x**^{2}**y**^{2}

Solution:

= x^{4} + y^{4 }– 2x^{2} y^{2} – 2x^{2} y^{2} – 7x^{2} y^{2}

= x^{4} + y^{4} + 2x^{2} y^{2} – 9x^{2} y^{2}

= (x^{2} + y^{2})^{2} – (3xy)^{2}

= (x^{2} + y^{2} + 3xy) (x^{2} + y^{2} – 3xy)

**(ii) 4x**^{4}**+ 25y**^{4}**+ 10x**^{2}**y**^{2}

Solution:

Take a = √2x ;b = √5y

a^{4} + b^{4} + a^{2} b^{2} = (a^{2} + b^{2} + ab) (a^{2} + b^{2} – ab)

4x^{4} + 25y^{4} + 10x^{2}y^{2} = (√2 x)^{4} + (√5 y)^{4} + 2 x^{2 }5 y^{2}

= [( 2 x^{2}) + ( 5 y^{2}) + (√10 xy] [( 2 x^{2}) + ( 5 y^{2}) – (√10 xy]

= (2x^{2} +5y^{2} + 10 xy) (2x^{2} +5y^{2} – 10 xy)

**(iii)9a ^{4}+100b^{4}+30a^{2}b^{2}**

Solution:

Take a = √3a ;b = √10b

a^{4} + b^{4} + a^{2} b^{2} = (a^{2} + b^{2} + ab) (a^{2} + b^{2} – ab)

9a^{4}+100b^{4}+30a^{2}b^{2} = (√3a)^{4}+(√10b)^{4}+3a^{2}.10b^{2}

= (3a^{2}+10b^{2}+30a^{2}b^{2}) (3a^{2}+10b^{2}– 30a^{2}b^{2})

**(iv)81a4 +9a2b2+b4**

Solution:

Take a = 3a ;b = b

a^{4} + b^{4} + a^{2} b^{2} = (a^{2} + b^{2} + ab) (a^{2} + b^{2} – ab)

81a^{4} +9a^{2}b^{2}+b^{4 }= (3a)^{4}+(3a)^{2}(b)^{2}+b^{4}

= (9a^{2}+b^{2}+3ab)(9a^{2}+b^{2}-3ab)

**(v) x**^{4}**– 6x**^{2}**y**^{2}**+ y**^{4}

= x^{4} + y^{4} + 2x^{2}y^{2} – 2x^{2}y^{2} – 6x^{2}y^{2}

= (x^{2} + y^{2})^{2} – ( 8 xy)^{2}

= (x^{2} + y^{2} + 8 xy) (x^{2} + y^{2} – 8 xy)

**vi) m**^{4}**+ n**^{4}**– 18m**^{2}**n**^{2}

= m^{4} + n^{4} + 2m^{2}n^{2} – 2m^{2}n^{2} – 18m^{2}n^{2}

= (m^{2} + n^{2})^{2} – 20 m^{2}n^{2}

= (m^{2} + n^{2} + √20 mn) (m^{2} + n^{2} – √20 mn)

**(vii) 4m**^{4}**+ 9n**^{4}**– 24m**^{2}**n**^{2}

= 4m^{4} + 9n^{4 }+ 12 m^{2}n^{2} – 12 m^{2}n^{2} – 24m^{2}n^{2}

= (2m^{2} + 3n^{2})^{2} – 36mn

= (2m^{2} + 3n^{2} + 6mn) (2m^{2} + 3n^{2} – 6mn)

**(viii) 9x**^{4 }**+ 4y**^{4}**+ 11x**^{2}**y**^{2 }

= 9x^{4 }+ 4y^{4} + 12x^{2}y^{2} – 12x^{2}y^{2} + 11x^{2}y^{2}

= (3x^{2})^{2} + (2y^{2})^{2} + 2 (3x)^{2} (2y)^{2} – x^{2}y^{2}

= (3x^{2} + 2y^{2})^{2} – x^{2}y^{2}

= (3x^{2} + 2y^{2 }+ xy) (3x^{2} + 2y^{2} – xy)

**Find the factors of the following**

**(i) x ^{4} + 9x^{2}+81**

Solution:

= x^{4}+9x^{2}+81+9x^{2}-9x^{2}

= (x^{2})^{2}+18x^{2}+81-9a^{2}

=(x^{2}+9)^{2}-(3a)^{2}

=( x^{2}+9 +3a)( x^{2}+9 – 3a)

**(ii) a**^{4}**+ 4a**^{2}**+ 16 **

Solution:

= a^{4} + 4a^{2} + 16 + 4a^{2 }– 4a^{2}

= (a^{2})^{2} + 8a^{2} + (4)^{2} – 4a^{2}

= (a^{2} + 4)^{2} – (2a)^{2}

= (a^{2} + 4 – 2a) (a^{2} + 4 + 2a)

**Factorise the following**

**(i) 64a**^{4}**+ 1 **

Solution:

Adding and subtracting 16a^{2} we get

64a4 + 1 + 16a^{2} – 16a^{2}

= (8a^{2})^{2} + 1 + 16a^{2} – 16a^{2}

= (8a^{2} + 1)^{2} – (4a)^{2}

= (8a^{2} + 1 + 4a) (8a^{2} + 1 – 4a)

**(ii) 3x**^{4}**+ 12y**^{4}

Solution:

3(x^{4} + 4y^{4})

3[(x^{2})^{2} + (^{2}y^{2})^{2}]

By adding and subtracting 2ab i. e.

2 x x^{2} x 2y = 4x^{2}y^{2}

= 3 [(x^{2})^{2} + 2(y^{2})^{2} + 4x^{2}y^{2} – 4x^{2}y^{2}]

= 3 [(x^{2} + 2y^{2})^{2} – (2xy)^{2}]

= 3 [(x^{2} + 2y2 + 2xy) (x^{2} + 2y^{2} – 2xy)]

**(iii) 4x**^{4}**+ 81y**^{4 }

Solution:

(2x^{2})^{2} + (9y^{2})^{2}

By adding and subtracting 2ab i. e.

2 x x^{2} x 9y^{2} = 36x^{2}y^{2 }

= (2x^{2})^{2} + (9y^{2})^{2} + 36x^{2}y^{2} – 36x^{2}y^{2 }

= (2x^{2} + 9y^{2})^{2} – (6xy)^{2}

= (2x^{2} + 9y^{2} + 6xy) (2x^{2} + 9y^{2} – 6xy)

**(iv) a**^{8}**– 16b**^{8}

Solution:

(a^{4})^{2} – (4b^{4})^{2}

Using a^{2} – b^{2} = (a + b) (a – b)

= (a^{4} + 4b^{4}) (a^{4} – 4b^{4})

= (a^{4} + 4b^{4}) [(a^{2})^{2} – (2b^{2})^{2}]

= (a^{4} + 4b^{4}) (a^{2} + 2b^{2}) (a^{2} – 2b^{2})

## 3.2.5 Some More Identities – Factorization[class 9]:

**Identity: a**^{3} + b^{3} = (a + b)(a^{2 }– ab + b^{2})

^{3}+ b

^{3}= (a + b)(a

^{2 }– ab + b

^{2})

We know that, (a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

Hence, a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b) = (a + b)[(a+b)^{2} – 3ab]

However,

(a + b)^{2} – 3ab = a^{2} + 2ab + b^{2} – 3ab = a^{2} – ab + b^{2}

Using this we obtain,

a^{3} + b^{3} = (a+b)(a^{2} – ab + b^{2})

**Identity 2: a**^{3} – b^{3} = (a – b)(a^{2 }+ ab + b^{2})

^{3}– b

^{3}= (a – b)(a

^{2 }+ ab + b

^{2})

Here we start with (a – b)^{3} = a^{3} – b^{3} – 3ab(a – b) . This gives, as earlier,

a^{3} – b^{3} = (a – b)^{3} + 3ab(a – b) = (a – b)[(a – b)^{2} + 3ab] = (a – b)(a^{2} + ab + b^{2})

Thus,

a^{3} – b^{3} = (a – b)(a^{2} + ab + b^{2})

**Example: Factorize x ^{3} + 27**

Solution:

We do this in several steps

**Step 1:** Write x^{3} + 27 in the form x^{3} + 3^{3} . This is in the form a^{3} + b^{3}.

**Step 2:** We substitute a = x and b = 3 in the factorization a^{3} + b^{3} = (a + b)(a^{2} – ab + b^{2})

We get,

x^{3} + 3^{3} = (x + 3)(x^{2} – 3x + 3^{2})

**Step 3:** Simplify the expression. This gives x^{3} + 27 = (x + 3)(x^{2} – 3x + 9)

## Factorization[class 9] – Exercise 3.2.5

**Factorise:**

**(i) 8y**^{3}**– 1 **

Solution:

= (2y)^{3} – 1^{3} [∵ a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})]

= (2y – 1) [(2y)^{2} + 2y .1 + 1^{2}]

= (2y – 1) (4y^{2} + 2y + 1)

**(ii) 27x**^{3}**– 8 **

Solution:

= (3x)^{3} – 2^{3 }[∵ a^{3 }– b^{3} = (a – b) (a^{2} + ab + b^{2})]

= (3x – 2) [(3x)^{2} + 3x .2 + 2^{2}]

= (3x – 2) (9x^{2} + 6x + 4)

**(iii) x**^{3 }**+ 8y**^{3}

Solution:

= x^{3} + (2y)^{3} [∵ a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})]

= (x + 2y) [x^{2} – x.2y + (2y)^{2}]

= (x + 2y) (x^{2 }– 2xy + 4y)^{2}

**(iv) 1 – x**^{3}

Solution:

= 1^{3} – x^{3} [∵a^{3 }– b^{3} = (a – b) (a^{2} + ab + b^{2})]

= (1 – x) (1^{2} + 1.x + x^{2})

= (1 – x) (1 + x + x^{2})

**(v) a**^{3}**b**^{3}**+ c**^{3}

Solution:

= (ab)^{3} + c^{3 }[∵ a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})]

= (ab + c) [(ab)^{2} – ab .c + c^{2}]

= (ab + c) (a^{2}b^{2} – abc + c^{2})

**(vi) a**^{3}**b – **^{𝐛}**/ _{𝟔𝟒} **

Solution:

= b (a^{3}– ^{1}/_{64})

= b [a^{3} – (^{1}/_{4} )^{3}]

= b (a – ^{1}/_{4}) [a^{2} + a. ^{1}/_{4} + ( ^{1}/_{4} )^{2}]

= b (a – 14) (a^{2} + ^{a}/_{4} + ^{1}/_{16} )

**(vii) ****𝐚**^{3}**/**_{𝟖}**+ 1 **

Solution:

= ( ^{a}/_{2} )^{3} + 1^{3} [∵ a^{3} + b^{3} = (a + b) (a^{2} – ab + b^{2})]

= ( ^{a}/_{2} + 1) [( ^{a}/_{2} )^{2} – ^{a}/_{2} .1 + 1^{2}]

= ( ^{a}/_{2} + 1) [a^{2}/_{4}− ^{a}/_{3}+ 1]

**(viii) 3a**^{6}**– ****𝐛**^{𝟔}**/**_{𝟗}

Solution:

= 3(a^{6}− b^{6}/27)

= 3[(a^{2})^{3}− (b^{2}/_{3})^{3}]

= 3[(a^{2}− b^{2}/_{3}) (a^{2})^{3}− a^{2}.b^{2}/_{3} (b^{2}/_{3})^{2}

= 3(a^{2}− b^{2}/_{3})( a^{4}+a^{2 }. b^{2}/_{3}+ b^{4}/_{3} )

**(ix) 2a**^{3}**+ **^{𝟏}**/ _{𝟒} **

Solution:

= 2 (a^{3} + ^{𝟏}/_{𝟒} )

= 2 (a^{3} + ^{𝟏}/_{2} )^{3}

= 2 (a + ^{𝟏}/_{2} ) [ a^{2} – a. ^{𝟏}/_{2} + (^{𝟏}/_{𝟒} )^{2}]

= 2 (a + ^{𝟏}/_{2} ) (a^{2} – ^{a}/_{2} + ^{𝟏}/_{𝟒} )

**(x) x**^{3}**– 512 **

Solution:

= x^{3} – 8^{3} [∵ a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})]

= (x – 8) (x^{2} + 8x + 64)

**(xi) 32x**^{3}**– 500 **

= 4 (8x^{3} – 125)

= 4 [(2x)^{3} – 5^{3}]

= 4 (2x – 5) [(2x)^{2} + 2x .5 – 5^{2}]

**= 4 (2x – 5) (4x**^{2}**+ 10 x + 25) **

**(xii) x**^{7}**+ xy**^{6 }

= x (x^{6} + y^{6})

= x [(x^{2})^{3} + (y^{2})^{3}]

= x [(x^{2} + y^{2}) { (x^{2})^{2} – x^{2} y^{2} + (y^{2})^{2}}]

**= x (x**^{2}**+ y**^{2}**) (x**^{4}**– x**^{2}**y**^{2}**+ y**^{2}**)**

** **

**(xiii) 2a**^{4 }**– 128a **

= 2a (a^{3} – 64)

= 2a (a^{3} – 4^{3})

= 2a (a – 4) (a^{2} + 4a + 4^{2})

**= 2a (a – 4) (a**^{2}**+ 4a + 16)**

**Factorise**

**(i) (1 – a)**^{3}**+ (3a)**^{3}

Solution:

Using x^{3} + y^{3} = (x + y) (x^{2} – xy + y^{2})

(1 – a)^{3} + (3a)^{3}

= (1 – a + 3a) [(1 – a)^{2} – (1 – a) 3a + (3a)^{2}]

= (1 + 2a) [(1 – a)^{2} – 3a + 3a^{2} + 9a^{2}]

= (1 + 2a) (1 + a^{2} – 2a – 3a + 12a^{2})

= (1 + 2a) (13a^{2} – 5a + 1)

**(ii) 8x**^{3}**– 27y**^{3}

Solution:

= (2x)^{3} – ( 3y)^{3 }

= ( 2x – 3y) [(2x)^{2} + 2x .3y + (3y)^{2}]

= (2x – 3y) (4x^{2} + 6xy + 9y^{2})

**(iii) z**^{4}**x**^{3}**+ 8y**^{3}**z**^{4}

Solution:

= z^{4 }(x^{3} + 8y^{3})

= z^{4} [x^{3} + (2y)^{3}]

= z^{4} (x + 2y) [x^{2} – x.2y + (2y)^{2}]

= z^{4} (x + 2y) [x^{2} – 2xy + 4y^{2}]

**(iv) 3(x + y)**^{3}**+ **^{𝟏}**/ _{9} **

**(xy)**

^{3}Solution:

= (x + y)^{3} + ^{1}/_{27} (xy)^{3}

= (x + y)^{3} + ( ^{xy}/_{3} )^{3}

= (x + y + ^{xy}/_{3} ) (x + y)^{2} – [ (x + y) ( ^{xy}/_{3} )+ ( ^{xy}/_{3})^{2}]

= (x + y + ^{xy}/_{3} ) (x^{2} + y^{2 }+ 2xy – (x + y)( ^{xy}/_{3}) + x^{2}y^{2}/_{9} )

**(v) x**^{6}**+ y**^{6}

Solution:

= (x^{2})^{3} – (y^{2})^{3}

= (x^{2} – y^{2}) [(x^{2})^{3 }+ x^{2} y^{2}– (y^{2})^{3}]

= (x^{2} – y^{2}) (x^{4} + x^{2} y^{2} + y^{4})

= (x + y) (x – y) (x^{2} + x y + y^{2}) (x^{2} – xy + y^{2})

** **

**(vi) a**^{3}**– 2√****𝟐 ****b**^{3}

Solution:

= a^{3} – (√2 b)^{3 }

= (a –√2b) [a^{2} + a.√2 b + √2 b^{2}]

= (a –√𝟐 b) (a^{2} + √𝟐 ab + √𝟐b^{2})

**Factorize the following**

**(i) x**^{6}**– 26x**^{3}**– 27**

Solution:

put x^{3} = a

a^{3} – 26a – 27

= a^{3} – 27a + a – 27

= a ( a – 27) + 1 (a – 27)

= (a – 27) ( a + 1)

= (x^{3} – 27) (x^{3} + 1)

= (x – 3) (x^{2} + 3x + 9) (x – 1) (x^{2} – x + 1)

**(ii) z**^{6}**– 63z**^{3}**– 64 **

Solution:

put z^{3} = a

a^{3} – 63a – 64

= a^{3} – 64a + a – 64

= a (a – 64) + 1 (a – 64)

= (a – 64) (a + 1)

= (z^{3} – 64) (z^{3} + 1)

= (z – 4) (z^{2} + 4z + 16) (z + 1) (z^{2} – z + 1)

**(iii) a**^{3}**– b**^{3}**– a + b **

Solution:

= (a – b) (a^{2} + ab + b^{2}) – (a – b)

= (a – b) [ a^{2} + ab + b^{2} – 1]

**(iv) x**^{6}**+ 7x**^{3}**– 8 **

Solution:

put x^{3} = a

a^{3} + 7a – 8

= a^{3} + 8a – a – 8

= a (a + 8) – 1 (a + 8)

= (a – 1) (a + 8)

= (x^{3} – 1) (x^{3} + 8)

= (x – 1) (x^{2 }+ x – 1) (x + 2) (x^{2} – 2x + 4)

**(v) a**^{3}**– **^{𝟏}**/ _{𝐚}**

^{𝟑}**– 2a +**

**𝟐𝐚**

Solution:

= (a – ^{𝟏}/_{𝐚}) (a^{2} + a.^{ 𝟏}/_{𝐚}+ ^{𝟏}/_{𝐚}^{2} ) – 2 (a – ^{𝟏}/_{𝐚} )

= (a – ^{𝟏}/_{𝐚} ) [a^{2 }+ 1 + ^{𝟏}/_{𝐚}^{2} – 2]

= (a – ^{𝟏}/_{𝐚} ) (a^{2} + ^{𝟏}/_{𝐚}^{𝟐} – 1 )

## 3.2.6 Factorization of some Cubic Expressions – Factorization[class 9]

- a
^{3}+ b^{3}+ c^{3 }– 3abc = (a + b + c)(a^{2}+ b^{2}+ c^{2}– ab – bc – ca)

We have,

a^{3} + b^{3} + c^{3 }– 3abc = (a^{3} + b^{3}) + c^{3} – 3abc

[use a^{3} + b^{3} = (a + b)^{3} – 3ab(a + b)]

= (a + b)^{3} + c^{3} – 3ab(a + b + c)

[use x^{3} + y^{3} = (x + y)(x^{2} – xy + y^{2})]

= (a + b + c)[(a + b)^{2} – (a + b)c + c^{2}] – 3ab(a + b + c)

= (a + b + c)[a^{2} + b^{2} + 2ab – ac – bc + c^{2} – 3ab]

=(a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)

**a ^{3} + b^{3} + c^{3 }– 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)**

**Corollary: If a + b + c = 0 then a ^{3} + b^{3} + c^{3} = 3abc**

Proof :

We use the identity

**a ^{3} + b^{3} + c^{3 }– 3abc = (a + b + c)(a^{2} + b^{2} + c^{2} – ab – bc – ca)**

Substituting a + b + c = 0, we obtain ,

a^{3} + b^{3} + c^{3} – 3abc = 0

Hence, a^{3} + b^{3} + c^{3} = 3abc

Such an identity is called a conditional identity. note that this is not valid for all choices of a, b, c it is valid only for those a, b, c satisfying a + b + c = 0

Alternatively, we can also do this as follows, Since a + b + c = 0 we get a + b = -c

Hence

(-c)^{3} = (a+ b)^{3} = a^{3} + b^{3} + 3ab(a+b) = a^{3} + b^{3} + 3ab(-c)

This simplifies to

a^{3} + b^{3} + c^{3} – 3abc = 0

or

a^{3} + b^{3} + c^{3} = 3abc

## Factorization[class 9] – Exercise 3.2.6

**Factorise**

**(i) a**^{3}**– b**^{3}**– c**^{3}**– 3abc**

Solution:

= a^{3} + (– b^{3})+(– c^{3}) – 3a(–b) ( –c)

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = a y = –b c = –c

∴ a^{3} + (– b^{3})+(– c^{3}) – 3a(–b) (c)

= [a + (–b) +(c)] [a^{2} + (–b)^{2} + (–c) – a(–b) (–b)(–c) – (–c)a]

= (a – b – c) (a^{2} + b^{2} + c^{2} + ab – bc + ca)

**(ii) a**^{3}**– b**^{3}**+ 8c**^{3}**+ 6abc **

Solution:

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = a y = –b c = 2c

a^{3} + (–b)^{3} + (2c)^{3} – 3a (–b) (2c)

= [a + (– b) + 2c] [a^{2} + (-b)^{2} + (2c)^{2} – a(– b) – (–b) (2c) – 2c.a]

= (a – b + 2c) (a^{2} + b^{2} + 4c^{2} + ab + 2bc – ca)

** **

**(iii) 125a**^{3}**+ b**^{3}**+ 64c**^{3}**– 60abc **

Solution:

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = 5a y = b c = 4c

(5a)^{3} + b^{3} + (4c)^{3} – 3(5a) b (4c)

= (5a + b + 4c) [(5a)^{2} + b^{2} + 16c^{2} – 5a(b) – b(4c) – 4(c) (5a)]

= (5a + b + 4c) (25a^{2} + b^{2}+ 16c^{2} – 5ab – 4bc – 20ca)

**(iv) 1 + b**^{3}**+ 8c**^{3}**– 6bc **

Solution:

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = 1 y = b c = 2c

1^{3} + b^{3} + (2c)^{3} – 3 . 1 . b . 2c

= (1 + b + 2c) [1^{2} + b^{2} + (2c)^{2} – 1 .b – b.2c – 2c.1]

= (1 + b + 2c) (1 + b^{2} + 4c^{2} – b – 2bc – 2c)

**(v) 8a**^{3}**+ 125b**^{3}**– 64c**^{3}**+ 120abc **

Solution:

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = 2a y =5b c = –4c

(2a)^{3} + (5b)^{3} + (–4c)^{3} – 3(2a) (5b) (-4c)

= [2a + 5b + (–4c)] [(2a)^{2} + (5b)^{2} + (4c)^{2} – (2a)(5b) – (5b)(-4c) –

(–4c) (2a)]

= (2a + 5b – 4c) (4a^{2} + 25b^{2} + 16c^{2} – 1ab + 20bc + 8ca)

**(vi) 2√****𝟐 ****a**^{3}**+ 16√****𝟐 ****b**^{3}**+ c**^{3}**– 12abc **

Solution:

Using x^{3} + y^{3} + z^{3} – 3xyz = (x + y +z) (x^{2} + y^{2} + z^{2} – xy – yz – xz) we get

x = √2a ;y =2√2 b; c = c

(√2 a)^{3} + (2 √2 b)^{3} + c^{3} – 3(√2 a) (2√2 b)c

= (√2 a + 2√2 b + c) [(√2 a)^{2} + (2√2 b)^{2} + c^{2} -√2 a. 2√2 b –

2√2 b . c – c. √2 a]

= (√𝟐 a + 2√𝟐 b + c) (2a^{2} + 8b^{2} + c^{2} – 4ab – 2√𝟐 bc – √𝟐 ac)

**(vii) (x – y)**^{3}**+ (y – z)**^{3}**+ (z – x)**^{3}

[hint: apply corollary]

Solution:

Using a + b + c =0 then a^{3} + b^{3} + c^{3} = 3abc using this

a = x – y b = y – z c = z – c

we have a + b + c = x – y + y – z + z – c = 0

∴ (x – y)^{3} + (y – z)^{3} + (z – x)^{3}

= ^{3} (x – y) (y – z) (z – x)

**(viii) p**^{3}**(q – r)**^{3}**+ q**^{3}**(r – p)**^{3}**+ (p – q)**^{3}

**[Hint: apply corollary] **

Solution:

Using identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} = ^{3}abc we get

a = p(q – r) b = q( r – p) c = r(p – q)

a + b + c = p(q – r) + q( r – p) + r(p – q)

= pq – pr – qr – pq + pr – qr

= 0

∴ p^{3} (q – r)^{3} + q^{3} (r – p)^{3} + (p – q)^{3}

= 3p (q – r) q(r – p) r(p – q)

= 3pqr (q – r) (r – p) (p – q)

**Find the product using appropriate identity**

**(i) (a – b – c) (a**^{2}**+ b**^{2}**+ c**^{2}**+ ab – bc – ca) **

Solution:

Using (x + y + z) (x^{2} + y^{2} + z^{2} – xy – yz – zx) = x^{3} + y^{3} + x^{3}

We get x = a y = –b z = –c

(a – b – c) (a^{2} + b^{2} + c^{2} + ab – bc – ca)

= a^{3} + (-b)^{3} + (-c)^{3}

= a^{3} – b^{3} – c^{3} – 3abc

**(ii) (x – 2y – z) (x**^{2}**+ 4y**^{2}**+ z**^{2}**+ 2xy – 2yz – zx) **

Solution:

Using (a + b + c) (a^{2} + b^{2} + c^{2 }– ab – bc ca) = a^{3} + b^{3} + c^{3} – 3abc

We get a = x b = –2y c = –z

[x + (–2y) + (–z)] [ x^{2} +(–2y)^{2} +(-z)^{2 }– x (-2y) – (-2y) (-z) + (-z)(x)]

= x^{3} + (–2y)^{3} + (-z)^{3} – 3x (–2y) (–8)

= x^{3} – 8y^{3 }– z^{3 }– 6xyz

**(iii) (x + y – z) (x**^{2}**+ y**^{2}**+ z**^{2}**– xy + yz + zx) **

Solution:

Using (a + b + c) (a^{2} + b^{2} + c^{2} – ab – bc ca) = a^{3} + b^{3} + c^{3} – 3abc

We get a = x b = y c = –z

[x + y + (–z)] [x^{2} + y^{2} + (–z)^{2} – xy + y(–z) + (–z) x]

= x^{3} + y^{3} + (-z)^{3} – 3(x) (y) (-z)

= x^{3} + y^{3} – z^{3} + 3xyz

**Get the factorization**

**(a + b + c)**^{3}**– a**^{3}**– b**^{3}**– c**^{3}**= 3 (a+ b) (b + c) (c + a) writing the expression **

**(a + b + c)**^{3}**– a**^{3}**– b**^{3}**– c**^{3}**= {(a + b + c)**^{3}**– a**^{3}**} – {b**^{3}**+ c**^{3}**} **

Solution:

We have to prove {(a + b + c)^{3} – a^{3}} – {b^{3} + c^{3}}

= 3(a + b) (b+ c) (b + c)

L.H.S = {(a + b + c)^{3} – a^{3}} – {b^{3} + c^{3}}

Using identity a^{3} – b^{3} = (a – b) (a^{2} + ab + b^{2})

= (a + b + c – a) {(a + b + c)^{2} + (a + b + c)a + a^{2}} – (b – c) (b^{2} – bc + c^{2})

= (b + c) [(a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ca + a^{2} + ab + ac + a^{2}) – (b^{2} – bc + c^{2})

= (b + c) [(3a^{2} + b^{2} + c^{2} + 3ab + 2bc + 3ca] – (b + c) (b^{2} – bc + c^{2})

= [3a^{2} + b^{2} + c^{2} + 3ab + 2bc + 3ca – b^{2} + bc – c^{2}]

= (b + c) (3a^{2} + 3ab + 3bc + 3ca)

= (b + c) 3[a (a + b) + c (a + b)]

= 3 (b + c) (a + b) (a + c)

= 3 (a + b) (a + c) (b + c)

= R.H.S

**If x + y + 4 = 0, find the value of x**^{3}**+ y**^{3}**– 12xy + 64.**

Solution:

x + y = – 4

Cubing on both sides

(x + y)^{3} = (– 4)^{3}

x^{3} + y^{3} + 3xy (x + y) = (– 4)^{3}

x^{3} + y^{3} + 3xy (– 4)^{3} = – 64

x^{3} + y^{3} – 12xy + 64 = 0

**If x = 2y + 6, find the value of x – 8y – 36xy – 216.**

Solution:

x – 2y = 6

Cubing on both sides

(x – 2y)^{3} = (6)^{3}

= x^{3} – (2y)^{3} – 3x (2y) (x – 2y) = 216

= x^{3} – 8y^{3} – 6xy (6) – 216 = 0

= x^{3} – 8y^{3} – 36xy – 216 = 0

**Without actually calculating the cubes, find the values of the following:**

**(i) (–12)**^{3}**+ 7**^{3}**+ 5**^{3}

Solution:

We have – 12 + 7 + 5 = 0

By the identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – 3abc we get

(–12)^{3} + 7^{3} + 5^{3} = 3 (–12) (7) (5)

= -1260

**(ii) (28)**^{3}**+ (–**^{3}**15)**^{3}**+ (–13)**^{3}

Solution:

We find that 28 – 15 – 13 = 0

By the identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – ^{3}abc we get

(28)^{3} + (–15)^{3} + (–13)^{3} = 3 x 28 x (–15) (–13)

= 16380

**(iii) (–15)**^{3}**+ 7**^{3}**+ 8**^{3}

Solution:

Since – 15 + 7 + 8 = 0

Using the identify if a + b + c = 0 then a^{3 }+ b^{3} + c^{3} – 3abc we get

(–15)^{3} + 7^{3} + 8^{3} = 3 (–15) 7 x 8

= – 2520

**(iv) (–10)**^{3}**+ 3**^{3}**+ 7**^{3}

Solution:

Since – 10 + 3 + 7 = 0

Using the identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – 3abc we get

(–10)^{3 }+ 3^{3} + 7^{3} = 3(-10) (3) (7)

= – 630

**Factorise the following using the identity**

**a**^{3}**+ b**^{3 }**+ c**^{3}**– 3abc = **^{𝟏}**/ _{𝟐} **

**(a + b + c) [(a – b)**

^{2}**+ (b – c)**

^{2 }**+ (c – a)**

^{2}**]**

**(i) ****a**^{3}**+ 8b**^{3}**+ c**^{3}**– 6abc **

Solution:

= ^{1}/_{2} (a + 2b + c) [(a – 2b)^{2} + (2b – c)^{2} + (c – a)^{2}]

= ^{1}/_{2} (a + 2b + c) (a^{2} + 4b^{2} – 4ab + 4b^{2} + c^{2} – 4bc + c^{2} +

a^{2} – 2ac)

= ^{1}/_{2} (a + 2b + c) (2a^{2} + 8b^{2} + 2c^{2} – 4ab – 4bc – 4ca)

**(ii) ****125a**^{3}**+ b**^{3}**+ 64c**^{3}**– 60abc **

Solution:

= ^{1}/_{2} (5a + b + 4c) [(5a – b)^{2} + (b – 4c)^{2} + (4c – 5a)^{2}]

= ^{1}/_{2} (5a + b + 4c) (25a^{2} + b^{2} – 10ab + b^{2 }+ 16c^{2} – 8bc + 16c^{2}

+ 25a^{2} – 40ac)

= ^{1}/_{2} (5a + b + 4c) (50a^{2 }+ 2b^{2 }+ 32c^{2} – 10ab – 8bc – 40ac)

** **

**(iii) ****1**^{3}**+ b**^{3}**+ 8c**^{3 }**– 6bc **

Solution:

= ^{1}/_{2} (1 + b + 2c) [(1 – b)^{2} + (b – 2c)^{2} + (2c – 1)^{2}]

= ^{1}/_{2} (1 + b +2c) (1 + b^{2} – 2b + b^{2} – 4bc + 4c^{2} + 4c^{2} + 1 – 4c)

= ^{1}/_{2} (1 + b + 2c) (2 + 2b^{2} + 8c^{2} – 2b – 4bc – 4c)

**(iv) ****125 – 8x**^{3}**– 27y**^{3}**– 90xy **

Solution:

= ^{1}/_{2} (5 – 2x – 3y) [(5 – 2x)^{2} + (–2x – 3y)^{2} + (–3y – 5)^{2}]

= ^{1}/_{2} (5 – 2x – 3y) (25 + 4x^{2} – 20x + 4x^{2} + 9y^{2} – 12xy + 9y^{2} +

25 – 30y)

= ^{1}/_{2} (5 –2x – 3y) (50 + 8x^{2} + 18y^{2} – 20x – 12xy – 30y)

**Factorise the following:**

**(i) (x – 2y)**^{3}**+ (2y – 3z)**^{3}**+ (3z – x)**^{3 }

Solution:

We find that x – 2y + 2y – 3z + 3z – x = 0

Hence using identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – 3abc we get

(x – 2y)^{3} + (2y – 3z)^{3} + (3z – x)^{3}

= 3(x – 2y) (2y – 3z) (3z – x)

**(ii) [****(𝐱**^{𝟐}**− 𝐲**^{𝟐}**)**^{𝟑}**+ (𝐲**^{𝟐}**− 𝐳**^{𝟐}**)**^{𝟑}**+(𝐳**^{𝟐}**− 𝐱**^{𝟐}**)**^{𝟑}**(𝐱 − 𝐲)**^{𝟑}**+ (𝐲 − 𝐳)**^{𝟑}**+(𝐳 − 𝐱)**^{𝟑}**]/[(x – y)**^{3}**(y – z)**^{3}**(z – x)**^{3}**] **

Solution:

We find that x^{2} – y^{2} + y^{2} – z^{2} + z^{2} – y^{2} = 0

and x – y + y – z + z – x = 0, using the identify if

a + b + c = 0 then a^{3} + b^{3} + c^{3} – ^{3}abc we get

[(x^{2}− y^{2})^{3}+ (y^{2}− z^{2})^{3}+(z^{2}− x^{2})^{3}] /[(x − y)^{3}+ (y − z)^{3}+(z − x)^{3}]

= [3(x^{2}− y^{2})+ (y^{2}− z^{2})+(z^{2}− x^{2})^{3}]/[(x – y) + (y – z)+(z – x)]

= [(x – y)(x + y)(y – z)(y + z)(z – x)(z + x)] /[(x – y)(y – z)(z – x)]

[∵ a^{2} – b^{2} = (a – b) (a + b)]

= (x + y) (y + z) (z + x)

**(iii) (2x – 3y)**^{3}**+ (4z – 2x)**^{3}**+ (3y – 4z)**^{3}

Solution:

We find that 2x – 3y + 4z – 2x + 3y – 4z = 0

Using the identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – 3abc we get

(2x – 3y)^{3} + (4z – 2x)^{3} + (3y – 4z)^{3}

= 3 (2x – 3y) (4z – 2x) (3y – 4z)

= 6 (2x – 3y) (4z – 2x) (3y – 4z)

**(iv) (a – 3b)**^{3}**+ (3b – c)**^{3}**+ (c – a)**^{3}

Solution:

We find that a – 3b + 3b – c + c – a = 0

Using the identify if a + b + c = 0 then a^{3} + b^{3} + c^{3} – 3abc we get

(a – 3b)^{3} + (3b – c)^{3} + (c – a)^{3}

= 3 (a – 3b) (3b – c) (c – a)

**Factorise the following:**

**(i) (x – y – z)**^{3}**– x**^{3}**+ y**^{3}**+ z**^{3}

Solution:

= [x + (–y) + (–z)]^{3} – x^{3} – (–y)^{3} – (–z)^{3}

Using the identify if (a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = x; b = –y; c = – z

= 3[x + (–y)] [(–y) + (–z)] [(–z) + x]

= 3(x – y) (–y –z) (–z + x)

= – 3(x – y) (y + z) (x – z)

= 3(x – y) (y + z) (z – x)

**(ii) (a – b – 1)**^{3}**– a**^{3}**+ b**^{3}**+ 1 **

Solution:

= [a + (–b)^{3} + (–1)^{3} – a^{3} – (–b)^{3} – (–1)^{3}

Using the identify if (x + y + z)^{3} – x^{3} – y^{3} – z^{3} = 3(x + y) (y +z) (z +x) we get

x = a y = –b z = – 1

= 3 [a + (–b)] [(–b) + (–1)] [(–1) + a]

= 3(a – b) (–b – 1) (–1 + a)

= 3 (a – b) (1 + b) (a – 1)

**(iii) (2x + y – z)**^{3}**– 8x**^{3}**– y**^{3 }**+ z**^{3 }

Solution:

= [2x + y + (–z)]^{3} – (2x)^{3 }– y^{3} – (–z)^{3 }

Using the identify if (a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = 2x b = y c = – z

= 3(2x + y) 2018 [(–z) + 2x]

= 3 (2x + y) (y – z) (-z + 2x)

= 3 (2x + y) (y – z) (2x – z)

**(iv) ( a – **^{𝟏}**/ _{𝟐} **

**b +**

^{𝟐}**/**

_{𝟑}**c )**

^{3}**– a**

^{3}**–**

^{𝟏}**/**

_{𝟖}**b**

^{3}**–**

^{𝟏}**/**

_{𝟐𝟕 }**c**

^{3}Solution:

= (a – (^{𝟏}/_{𝟐} b) + ^{𝟐}/_{𝟑} c )3 – a^{3} – ( ^{1}/_{8} b)3 – ^{𝟏}/_{𝟐𝟕} c3

Using the identify if (x + y + z)3 – x3 – y3 – z3 = 3(x + y) (y +z) (z +x) we get

x = a ;y = – ^{b}/_{𝟐}; z = – ^{c}/_{𝟑}

= 3 (a – ^{b}/_{𝟐} ) ( −b2 + ^{c}/_{3} ) ( ^{c}/_{3} – a)

= 3 (a – ^{b}/_{𝟐} ) ( ^{c}/_{3} – ^{b}/_{2}) (a + ^{c}/_{3} )

**(v) (x + y + z)**^{3}**– (2x – y)**^{3}**– (2y – z)**^{3}**– (2z – x)**^{3}

Solution:

Let us consider 2x – y + 2y – z + 2z – x = (x + y + z)

∴ The given problem can be written as

(2x – y + 2y – z + 2z – x)^{3} – (2x – y)^{3} – (2y – z)^{3} – (2z – x)^{3}

Using the identify

(a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = 2x – y; b = 2y – z; c = 2z – x

= 3{(2x – y) + (2y – z)}{(2y – z) + (2z – x)}{(2z – x) + (2x – y)}

= 3 (2x – y + 2y – z) (2y – z + 2z – x) (2z – x + 2x – y)

= 3 (2x + y – z) (2y + z – x) (2z + x – y)

**(vi) (x + y + z – 3)**^{3}**– (x – 1)**^{3}**– (y – 1)**^{3}**– (z – 1)**^{3}

Solution:

= [(x – 1) + (y – 1) + (z – 1)]^{3} – (x – 1)^{3} – (y – 1)^{3} – (z – 1)^{3}

Using the identify

(a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = x – 1; b = y – 1; c = z – 1

= 3 {x – 1 (y – 1)} + {y – 1 + (z – 1)} {z – 1 + (x – 1)}

= 3 (x – 1 + y – 1) (y – 1+ z – 1) (z – 1 + x – 1)

= 3 (x + y – z) (y + z – 2) (z + x – 2)

**Find the factors of the following numbers**

**(i) 30**^{3}**– 12**^{3}**– 10**^{3}**– 8**^{3}

Solution:

We find that 12 + 10 + 8 = 30

Using the identify

(a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = 12; b = 10 ;c = 8

= 3 (12 + 10) (10 + 8) (8 + 12)

= 3 x 22 x 18 x 20

= 2^{4} x 3^{3}. 5.11

**(ii) 85**^{3}**– 68**^{3}**+ 5**^{3}**– 22**^{3}

Solution:

We find that 68 – 5 + 22 = 85

Using the identify

(a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = 68 b = –5 c = 22

= 3 {68 + (–5)} {–5 + 22} {22 + 68} = 3 x 63 x 17 x 90

= 2.3^{5}.5.7.17

**(iii) 100**^{3}**– 49**^{3}**+ 10**^{3 }**– 61**^{3}

Solution:

We find that 49 – 10 + 61 = 100

Using the identify

(a + b + c)^{3} – a^{3} – b^{3} – c^{3} = 3(a + b) (b +c) (c +a) we get

a = 49; b = –10; c = 61

= 3 [49 + (10)] (–10 + 61) (61 + 49)

= 3 x 39 x 51 x 110

= 2.3^{3}.5.11.13.17

**Prove that**

**(x + y + z)**^{3}**– (x + y – z)**^{3 }**– (y + z – x)**^{3 }**– (z + x – y)**^{3}**= 24xyz **

**[Hint: find a relation between x + y + z and x + y – z, z + x – y] **

Solution:

L.H.S. (x + y + z)^{3} – (x + y – z)^{3} – (y + z – x)^{3} – (z + x – y)^{3}

= (x + y + z)^{3 }– (x + y – z)^{3} – [(y + z – x)^{3} + (z + x – y)^{3}]

= [x + y + z – (x + y – z)] [(x + y + z)^{2} + (x + y + z) (x + y – z)] –

(x + y – z)^{2}] – (y + z – x + z + x – y)] [(y + z – x)^{2} – (y + z – x)

(z + x – z) + (z + x –y)^{2}]

= [2z {y^{2} + z^{2} + x^{2} + 2yz – 2zx – 2xy – (yz + z^{2} – zx + xy + zx – x^{2} – y^{2}

– yz + xy) + z + x + y + 2xy – 2xy – 2yz}]

= [2z (3x^{2} + 3y^{2} + z^{2} + 6xy) – 2z (y^{2} + z^{2} + x^{2} + 2yz – 2zx – 2xy – yz – z^{2}

+ zx – xy – zx + x^{2} + y^{2} + yz – xy + z^{2} + x^{2} + y^{2} + 2xy – 2xy – 2yz)

= (x^{2}z + 6y^{2}z + 2z^{3} + (x + y) – 2z (3x^{2} + 3y^{2} – 6xy)

= 6x^{2}z + 6y^{2}z + 12xyz – 6x^{2}z – 6y^{2}z – 2z^{3} + 12xyz

= 24xyz

= R.H.S