Factorization[class 9] – Full Chapter Exercise solutions

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 9x2 + 12xy

Solution:

We write 9x2 + 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 9x2 + 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

  1. Factorise:

(i) 5x2 – 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) 18x2y – 24xyz

Solution:

= 6xy (3x – 4z)


(v) 7xa – 70 xb

Solution:

= 7x (a – 10b)


(vi) 13m2 + 156n2

Solution:

=13(m2 + 12n2)


(vii) 3x2 + 6x +6

Solution:

= 3 (x2 + 2x + 2)


(viii) 4abx2 + 8abx + 120by

Solution:

= 4ab (x2 + 2x +3y)


2. Factorise

(i) x2 + xy + xz + yz

Solution:

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

=(x + y) (x + z)


(ii) a2 – ab + ac – bx

Solution:

= a (a – b) + c (a – b)

= (a – b) (a – c)


(iii) 4x + bx + 4b + b2

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) (x2 + 1)


(v) a3 + a2b + ab + b2

Solution:

= a2 (a +b) + b (a +b)

= (a + b) (a2 + b)


(vi) 3x3 + 5x2 + 3x +5

Solution:

= x2 (3x +5) + 1 (3x + 5)

= (3x + 5) (x2 + 1)


(vii) y4 – 2y3 + y – 2

Solution:

= y3(y – 2) + 1 (y – 2)

= (y – 2) (y3 – 1)

= (y – 2) (y3 – 13)

= (y – 2) (y + 1) (y2 – y +1)


(viii) t4 + t4 – 2t – 2

Solution:

= t3 (t + 1) – 2(t +1)

= (t – 1) (t3 – 2)


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

We know that (a + b)(a – b) = a2 – b2. We can use this identity to factorise many expressions involving the difference of the two squares.

Example 8: Factorise 25x2 – 64y2

Solution:

We write 25x2 – 64y2

25x2 – 64y2 = (5x)2 – (8y)2

This is in the form a2 – b2 which factorises to (a – b)(a + b). Thus,

25x2 – 64y2 = (5x)2 – (8y)2 = (5x – 8y)(5x + 8y)


Factorization[class 9] – Exercise 3.2.1

  1. Factorize the following:

(i) 9x2 – 16y2

Solution:

= (3x)2 – (4y)2

Using a2 – b2 = (a + b) (a – b)

= (3x + 4y) (3x – 4y)


(ii) 5x2 – 7y2 [ Hint : ( 𝟓 )2 = 5 ]

Solution:

( 5 x)2 – ( 7 y)2

Using a2 – b2 = (a + b) (a – b)

= ( 5 x + 7 y) ( 5 x – 7 y)


(iii) 100 – 9x2

Solution:

= 102 – (3x)2

= (10 + 3x) (10 – 3x) [∵a2 – b2 = (a + b) (a – b)]


(iv) (x + 4y)2 – 4z2

Solution:

= (x + 4y)2 – (2z)2 [∵a2 – b2 = (a + b) (a – b)]

= (x + 4y + 2z) (x + 4y – 2z)


(v) x – 64xy4

Solution:

= x (1 – 64y4)

= x (1 + (8y2)2) [∵a2 – b2 = (a + b) (a – b)]

= x (1 + 8y2) (1 – 8y2)


(vi) a2 + 2ab + b2 – 9c2

Solution:

= (a + b)2 – (3c)2 [∵(a + b)2 = a2 + 2ab + b2 ;a2 – b2 = (a + b) (a – b)]

= (a + b + 3c) (a + b – 3c)


(vii) 4x2 – 9y2 – 2x – 3y

Solution:

= (2x)2 – (3y)2 – (2x + 3y)

= [(2x + 3y) (2x – 3y)] – (2x – 3y) [∵a2 – b2 = (a + b) (a – b)]

= (2x + 3y) (2x – 3y – 1)


(viii) a2 + b2 – a + b

Solution:

= a2 + b2 – (a – b) [∵a2 – b2 = (a + b) (a – b)]

= (a + b) (a – b) – (a – b)

= (a – b) (a + b – 1)


(ix) x4 – 625

Solution:

= (x2)2 – 252 [∵a2 – b2 = (a + b) (a – b)]

= (x + 25) (x2 – 25)

= (x2 + 25) (x2 – 52)

= (x2 + 25) (x + 5) (x – 5)


(x) 36 (5x + y)2 – 25 (4x – y)2

Solution:

= [6 (5x + y)]2 – [5 (4x – y)]2 [∵a2 – b2 = (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) 12m2 – 75n2

Solution:

= ( √12 m2)2 – (√75 n2)2 [∵a2 – b2 = (a + b) (a – b)]

= (2 √3m2)2 – (5√3 n2)2 [ 12 =2√3 and 75 = 5√3

= (2√3m2 + 5√3 n2) (2√3 m2 – 5√3 n2)

= √3 (2m2 + 5n2) (2m – 5n) (2m + 5n)


  1. Factorize by adding and subtracting appropriate quantity.

(i) x2 + 6x + 3

Solution:

= x2 + 2. x. 3 + 32 – 32 + 3

= (x + 3)2 – 9 + 3

= (x + 3)2 – 6 [∵a2 – b2 = (a + b) (a – b)]

= (x + 3)2 – 62

= (x + 3 + 6 ) (x + 3 – 6 )


(ii) x2 + 10x + 8

Solution:

= x2 + 2. x. 5 + 52 – 52 + 8

= (x + 5)2 – 25 + 8

= (x + 5)2 – 17 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 172

= (x + 5 + 17 ) (x + 5 – 17 )


(iii) x2 + 10x + 20

Solution:

= x2 + 2. x. 5 + 52 – 52 + 20 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 5)2 – 25 + 20

= (x + 5)2 – 5

= (x + 5)2 – 52 [∵a2 – b2 = (a + b) (a – b)]

= (x + 5 + 5 ) (x + 5 – 5 )


(iv) x2 + 2x – 1

Solution:

= x2 + 2. x. 1 + 12 – 12 + 1 [∵a2 + 2ab + b2 = (a + b)2]

= (x + 1)2 – 1 + 1

= (x + 1)2 – 2

= (x + 1)2 – 22 [∵a2 – b2 = (a + b) (a – b)]

= (x + 1 + 2 ) (x + 1 – 2 )


  1. prove that (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖 = 1

Solution:

L.H.S = (1 + 𝐱/𝐲 ) (1 – 𝐱/𝐲 ) (1 + 𝐱𝟐/𝐲𝟐) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

[∵a2 – b2 = (a + b) (a – b)]

= (1 – 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟐/𝐲𝟐 ) (1 + 𝐱𝟒/𝐲𝟒) + 𝐱𝟖/𝐲𝟖

= (1 – 𝐱𝟒/𝐲𝟒) (1 + 𝐱𝟒/𝐲𝟒 ) + 𝐱𝟖/𝐲𝟖

= (1)2 – (𝐱𝟒/𝐲𝟒)2 + 𝐱𝟖/𝐲𝟖

= 1 – 𝐱𝟖/𝐲𝟖 + 𝐱𝟖/𝐲𝟖

= 1 R.H.S


  1. If x = 𝐚+𝐛/𝐚−𝐛 and y = 𝐚−𝐛/𝐚+𝐛 find x2 – y2.

Solution:

x2 – y2 = (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 ]/(a2− b2)][[(a+b)2− (a−b)2]/(a2− b2)]

= {(a2+ 2ab + b2+ a2+ b2− 2ab)/[(a2− b2)2]} x (a2 + 2ab + b2 – a2 – b2 + 2ab)

= [(2a2+ 2b2)4ab]/[(a2− b2)2]

= [2(a2+ b2)4ab]/(a2− b2)2

= [8ab(a2+ b2)]/(a2− b2)2


  1. If p = x – y, q = y – z and r = z – x, simplify r2 – p + 2pq – q2

Solution:

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

= z2 – 2xy + x2 – (x2 – 2xy + y2) + 2 (xy – y2 – xz + yz) – (y2 + z2 – 2yz)

= z2 + x2 – 2xz – x2 + 2xy – y2 + 2xy – 2y2 – 2xz + 2yz – y2 – z2 + 2yz

= 4xy – 4xz + 4yz – 4y2

= 4[xy – xz + yz – y2]

= 4[x (y – z) – y (y – z)]

= 4 (x – y) (y – z)

= 4pq


  1. 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:

Factorization[class 9]

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

OA2 = OB2 + AB2

132 = x2 + 52

x2 = 132 – 52

= (13 + 5) (13 – 5)

= 18 (8)

x2 = 144 (∵ the perpendicular line drawn from the centre of the circle to a chord bisects the chord)

x = 144

x = 12 cm


  1. Is it possible to factorize x2 + 4x + 20? Give reason

Solution:

x2 + 4x + 20

= x2 + 2. x. 2 + 22 – 22 + 20

= x2 + 4x + 4 – 4 + 20

= (x + 4)2 + 16

= (x + 4)2 + 42

We cannot factorize their further since this is nit of the form a2 – b2


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

Example: Factorize x2 + 4√2x + 8

Solution:

We write this in the form

x2 + 2(x)(2√2) + (2√2)2

This is in the form a2 + 2ab + b2 , where a = x, b = 2√2

But we know that a2 + 2ab + b2 = (a + b)2

Thus,

x2 + 4√2x + 8 = (x + 2√2)2

Example: Factorise x2 + 11x + 30

Solution:

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

This helps us to split the trinomial as,

x2 + 11x + 30 = x2 + 5x + 6x + 30

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

= (x + 6)(x + 5)

Thus (x+ 5) and (x + 6) are the factors of x2 + 11x + 30

Example: Factorise (x2 – 2x)2 – 23(x2 – 2x) + 120

Solution:

We introduce a = x2 – 2x . Then the expression is a2 – 23a + 120. We observe that – 23 = (-15) + (-8) and 120 = (-15)(-8). Hence we can write a2 – 23a + 120 = a2 – 15a – 8a + 120 = a(a – 15) – 8(a – 15) = (a – 15)(a – 8)

Hence, (x2 – 2x)2 – 23(x2 – 2x) + 120 = (x2 – 2x – 15)( x2 – 2x – 8)

Further we see,

(x2 – 2x – 15) = x2 – 5x + 3x – 15 = x(x – 5)+3(x -5) = (x + 3)(x – 5)

( x2 – 2x – 8) = x2 – 4x + 2x – 8 = x(x – 4)+2(x – 4) = (x – 4)(x + 2)

We hence obtain

(x2 – 2x)2 – 23(x2 – 2x) + 120 = (x + 3)(x – 5)(x + 2)(x – 4)


Factorization[class 9] – Exercise 3.2.3

  1. Factorize

(i) x2 + 9x + 18

Solution:

= x2 + 6x + 3x + 18

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

= (x + 6) (x + 3)

 

(ii) y2 + 5y – 24

Solution:

= y2 + 8y – 3y – 24

= y (y + 8) – 3 (y + 8)

= (y + 8) (y – 3)

 

(iii) 7y2 + 49y +84

Solution:

= 7 (y2 + 7y + 12)

= 7 (y2 + 4y + 3y + 12)

=7 2017

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

 

(iv) 40 + 3x – x2

Solution:

= – [x2 – 3x – 40]

= 3 – [x2 – 8x + 5x – 40]

= – [x (x – 8) + 5 (x – 8)]

= [(x – 8) (x + 5)]

= (8 – x) (x + 5)

 

(v) m2 + 17mn – 84n2

Solution:

= m2 + 21mn – 4mn – 84n2

= m (m + 21n) – 4n (m + 21n)

= (m + 21m) (m – 4n)

 

(vi) 117p2 + 2pq – 24q2

Solution:

= 117p2 + 54pq – 52pq – 24q2

= 9p (13p + 6q) – 3q (13p + 6q)

= (9p – 3q) (13p + 6q)

 

(vii) 15x2 – r – 28

Solution:

= 15x2 – 21x + 20x – 28

= 3x (5x – 7) + 4 (5x – 7)

= (3x + 4) (5x – 7)

 

(viii) 2x2 – x – 21

Solution:

= 2x2 – 7x + 6x – 21

= x (2x – 7) + 3 (2x – 7)

= (2x – 7) (x + 3)

 

(ix) 8k2 – 22k – 21

Solution:

= 8k2 – 28k + 6k – 21

= 4k (2k – 7) + 3(2k – 7)

= (2k – 7) (4k + 3)

 

(x) 𝟏/𝟑 x2 – 2x – 9

Solution:

= (x2+ 6x−27)/3

= 1/3 [x2 – 6x – 27]

= 1/3 [x2 – 9x + 3x – 27]

= 1/3 [x (x – 9) + 3 (x – 9)]

=1/3 (x – 9) (x + 3)


  1. Factorize

(i) √𝟓 x2 + 2x – 3√𝟓

Solution:

= √5 x2 + 5x – 3x – 3√5

= √5 x (x + √5 ) – 3 ((x +√ 5 )

= (x + √5 ) (√5 x – 3)

 

(ii) √𝟑 a2 + 2a – 5√𝟑

Solution:

= √3a2 + 5a – 3a – 5√3 – 5√3 – √3

= a (√3 a + 5) – √3 (√3 a + 5)

= (√3 a + 5) (a – √3 )

 

(iii) 7√2 y2 – 10y – 4√2

Solution:

= 7√2 y2 – 14y + 4y – 4√2

= 7√2 y (y – √2) + 4 (y – 2)

= (y – √2) (7√2 y + 4)

 

(iv) 6√𝟑 z2 – 47z – 5√𝟑

Solution:

= 6√3 z2 – 45z – 2z – 5√3

= 3√3 z (2z –5√3) – (2z – 5√3)

= (2z – 5√3) (3√3z –1)

 

(v) 4√𝟑x2 + 5x – 2√𝟑

Solution:

= 4√3 x2 + 8x – 3x – 2√3

= 4x (√3x + 2) – √3(√3x + 2)

= (√3x + 2) (4x –√3)


  1. Factorize

(i) 2 (x + y)2 – 9 (x + y) – 5

Solution:

Let x + y = p

= 2p2 – 9p – 5

= 2p2 – 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

= 2x2 – 25x + 12

= 2x2 – 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

= 12a2 – 25ab + 12b2

= 12a2 – 16ab – 9ab + 12b2

= 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)


  1. Factorize

(i) x4 – 3x2 + 2

Solution:

Put x2 = a

a2 – 3a + 2

= a2 – 2a – a + 2

= a (a – 2) – 1 (a – 2)

= (a – 2) (a – 1)

= (x2 – 2) (x2 – 1)

= (x – 2 ) (x + 2 ) (x – 1) (x + 1) [∵a2 – b2 = (a + b) (a – b)]


(ii) 4x4 + 7x2 – 2

Solution:

Put x2 = a

= 4a2 + 7a – 2

= 4a2 + 8a – a – 2

= 4a (a + 2) – 1 (a + 2)

= (a + 2) (4a – 1) [∵a2 – b2 = (a + b) (a – b)]

= (x2 + 2) (4x2 – 1)

= (x2 + 2) (2x + 1) (2x – 1)


(iii) 3x3 – x2 – 10x

Solution:

= x (3x2 – x – 10)

= x (3x2 – 6x + 5x – 10)

= x [3x (x – 2) + 5 (x – 2)]

= x (x – 2) (3x + 5)


(iv) 8x3 + 2x2y – 15xy2

Solution:

= x (8x2 – 2xy – 15y2)

= x (8x2 – 12xy + 10xy – 15y2)

= x [4x (2x – 3y) + 5y (2x – 3y)]

= x (2x – 3y) (4x + 5y)


(v) x6 – 7x3 – 8

Solution:

Put x3 = a

a3 – 7a – 8 [∵a3 + b3 = (a + b) (a2 – ab + b2)]

= a3 – 8a + a – 8 [∵a3 – b3 = (a – b) (a2 – ab + b2)]

= a (a – 8) + 1 (a – 8)

= (a – 8) (a + 1)

= (x3 – 3) (x3 + 1)

= (x3 – 3) (x3 + 1)

= (x + 1) (x2 – x + 1) (x – 2) (x2 + 2x + 4)


3.2.4 Some Miscellaneous Factorization – Factorization[class 9]:

Consider the expression a4 + a2b2 + b4 .

a4 + a2b2 + b4 = (a2 + b2 + ab)(a2 + b2 – ab)

Example: Factorize 4x4 + 9y4 + 6x2y2

Solution:

a = √2x , b = √3y we obtain,

a4 + a2b2 + b4 = 4x4 + 6x2y2 + 9y4

We can use the factorization a4 + a2b2 + b4 = (a2 + b2 + ab)(a2 + b2 – ab)

4x4 + 6x2y2 + 9y4 = (2x2 – √6xy + 3y2)(2x2 + √6xy + 3y2)


Factorization[class 9] – Exercise 3.2.4

  1. Resolve into factors

(i) x4 + y4 – 7x2 y2

Solution:

= x4 + y4 – 2x2 y2 – 2x2 y2 – 7x2 y2

= x4 + y4 + 2x2 y2 – 9x2 y2

= (x2 + y2)2 – (3xy)2

= (x2 + y2 + 3xy) (x2 + y2 – 3xy)

 

(ii) 4x4 + 25y4 + 10x2y2

Solution:

Take a = √2x ;b = √5y

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

4x4 + 25y4 + 10x2y2 = (√2 x)4 + (√5 y)4 + 2 x2 5 y2

= [( 2 x2) + ( 5 y2) + (√10 xy] [( 2 x2) + ( 5 y2) – (√10 xy]

= (2x2 +5y2 + 10 xy) (2x2 +5y2 – 10 xy)

(iii)9a4+100b4+30a2b2

Solution:

Take a = √3a ;b = √10b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

9a4+100b4+30a2b2 = (√3a)4+(√10b)4+3a2.10b2

= (3a2+10b2+30a2b2) (3a2+10b2– 30a2b2)

(iv)81a4 +9a2b2+b4

Solution:

Take a = 3a ;b = b

a4 + b4 + a2 b2 = (a2 + b2 + ab) (a2 + b2 – ab)

81a4 +9a2b2+b4 = (3a)4+(3a)2(b)2+b4

= (9a2+b2+3ab)(9a2+b2-3ab)

 

(v) x4 – 6x2y2 + y4

= x4 + y4 + 2x2y2 – 2x2y2 – 6x2y2

= (x2 + y2)2 – ( 8 xy)2

= (x2 + y2 + 8 xy) (x2 + y2 – 8 xy)

 

  1. vi) m4 + n4 – 18m2n2

= m4 + n4 + 2m2n2 – 2m2n2 – 18m2n2

= (m2 + n2)2 – 20 m2n2

= (m2 + n2 + √20 mn) (m2 + n2 – √20 mn)

 

(vii) 4m4 + 9n4 – 24m2n2

= 4m4 + 9n4 + 12 m2n2 – 12 m2n2 – 24m2n2

= (2m2 + 3n2)2 – 36mn

= (2m2 + 3n2 + 6mn) (2m2 + 3n2 – 6mn)

 

(viii) 9x4 + 4y4 + 11x2y2

= 9x4 + 4y4 + 12x2y2 – 12x2y2 + 11x2y2

= (3x2)2 + (2y2)2 + 2 (3x)2 (2y)2 – x2y2

= (3x2 + 2y2)2 – x2y2

= (3x2 + 2y2 + xy) (3x2 + 2y2 – xy)


  1. Find the factors of the following

(i) x4 + 9x2+81

Solution:

= x4+9x2+81+9x2-9x2

= (x2)2+18x2+81-9a2

=(x2+9)2-(3a)2

=( x2+9 +3a)( x2+9 – 3a)

 

(ii) a4 + 4a2 + 16

Solution:

= a4 + 4a2 + 16 + 4a2 – 4a2

= (a2)2 + 8a2 + (4)2 – 4a2

= (a2 + 4)2 – (2a)2

= (a2 + 4 – 2a) (a2 + 4 + 2a)


  1. Factorise the following

(i) 64a4 + 1

Solution:

Adding and subtracting 16a2 we get

64a4 + 1 + 16a2 – 16a2

= (8a2)2 + 1 + 16a2 – 16a2

= (8a2 + 1)2 – (4a)2

= (8a2 + 1 + 4a) (8a2 + 1 – 4a)

 

(ii) 3x4 + 12y4

Solution:

3(x4 + 4y4)

3[(x2)2 + (2y2)2]

By adding and subtracting 2ab i. e.

2 x x2 x 2y = 4x2y2

= 3 [(x2)2 + 2(y2)2 + 4x2y2 – 4x2y2]

= 3 [(x2 + 2y2)2 – (2xy)2]

= 3 [(x2 + 2y2 + 2xy) (x2 + 2y2 – 2xy)]

 

(iii) 4x4 + 81y4

Solution:

(2x2)2 + (9y2)2

By adding and subtracting 2ab i. e.

2 x x2 x 9y2 = 36x2y2

= (2x2)2 + (9y2)2 + 36x2y2 – 36x2y2

= (2x2 + 9y2)2 – (6xy)2

= (2x2 + 9y2 + 6xy) (2x2 + 9y2 – 6xy)

 

(iv) a8 – 16b8

Solution:

(a4)2 – (4b4)2

Using a2 – b2 = (a + b) (a – b)

= (a4 + 4b4) (a4 – 4b4)

= (a4 + 4b4) [(a2)2 – (2b2)2]

= (a4 + 4b4) (a2 + 2b2) (a2 – 2b2)


3.2.5 Some More Identities – Factorization[class 9]:

Identity: a3 + b3 = (a + b)(a2 – ab + b2)

We know that, (a + b)3 = a3 + b3 + 3ab(a + b)

Hence, a3 + b3 = (a + b)3 – 3ab(a + b) = (a + b)[(a+b)2 – 3ab]

However,

(a + b)2 – 3ab = a2 + 2ab + b2 – 3ab = a2 – ab + b2

Using this we obtain,

a3 + b3 = (a+b)(a2 – ab + b2)

Identity 2: a3 – b3 = (a – b)(a2 + ab + b2)

Here we start with (a – b)3 = a3 – b3 – 3ab(a – b) . This gives, as earlier,

a3 – b3  = (a – b)3 + 3ab(a – b) = (a – b)[(a – b)2 + 3ab] = (a – b)(a2 + ab + b2)

Thus,

a3 – b3  = (a – b)(a2 + ab + b2)

Example: Factorize x3 + 27

Solution:

We do this in several steps

Step 1: Write x3 + 27 in the form x3 + 33 . This is in the form a3 + b3.

Step 2: We substitute a = x and b = 3 in the factorization a3 + b3 = (a + b)(a2 – ab + b2)

We get,

x3 + 33 = (x + 3)(x2 – 3x + 32)

Step 3: Simplify the expression. This gives x3 + 27 = (x + 3)(x2 – 3x + 9)


Factorization[class 9] – Exercise 3.2.5

  1. Factorise:

(i) 8y3 – 1

Solution:

= (2y)3 – 13 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (2y – 1) [(2y)2 + 2y .1 + 12]

= (2y – 1) (4y2 + 2y + 1)

 

(ii) 27x3 – 8

Solution:

= (3x)3 – 23 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (3x – 2) [(3x)2 + 3x .2 + 22]

= (3x – 2) (9x2 + 6x + 4)

 

(iii) x3 + 8y3

Solution:

= x3 + (2y)3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (x + 2y) [x2 – x.2y + (2y)2]

= (x + 2y) (x2 – 2xy + 4y)2

 

(iv) 1 – x3

Solution:

= 13 – x3 [∵a3 – b3 = (a – b) (a2 + ab + b2)]

= (1 – x) (12 + 1.x + x2)

= (1 – x) (1 + x + x2)

 

(v) a3 b3 + c3

Solution:

= (ab)3 + c3 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= (ab + c) [(ab)2 – ab .c + c2]

= (ab + c) (a2b2 – abc + c2)

 

(vi) a3b – 𝐛/𝟔𝟒

Solution:

= b (a31/64)

= b [a3 – (1/4 )3]

= b (a – 1/4) [a2 + a. 1/4 + ( 1/4 )2]

= b (a – 14) (a2 + a/4 + 1/16 )

 

(vii) 𝐚3/𝟖 + 1

Solution:

= ( a/2 )3 + 13 [∵ a3 + b3 = (a + b) (a2 – ab + b2)]

= ( a/2 + 1) [( a/2 )2a/2 .1 + 12]

= ( a/2 + 1) [a2/4a/3+ 1]

 

(viii) 3a6 𝐛𝟔/𝟗

Solution:

= 3(a6− b6/27)

= 3[(a2)3− (b2/3)3]

= 3[(a2− b2/3) (a2)3− a2.b2/3 (b2/3)2

= 3(a2− b2/3)( a4+a2 . b2/3+ b4/3 )

 

(ix) 2a3 + 𝟏/𝟒

Solution:

= 2 (a3 + 𝟏/𝟒 )

= 2 (a3 + 𝟏/2 )3

= 2 (a + 𝟏/2 ) [ a2 – a. 𝟏/2 + (𝟏/𝟒 )2]

= 2 (a + 𝟏/2 ) (a2a/2 + 𝟏/𝟒 )

 

(x) x3 – 512

Solution:

= x3 – 83 [∵ a3 – b3 = (a – b) (a2 + ab + b2)]

= (x – 8) (x2 + 8x + 64)

 

(xi) 32x3 – 500

= 4 (8x3 – 125)

= 4 [(2x)3 – 53]

= 4 (2x – 5) [(2x)2 + 2x .5 – 52]

= 4 (2x – 5) (4x2 + 10 x + 25)

 

(xii) x7 + xy6

= x (x6 + y6)

= x [(x2)3 + (y2)3]

= x [(x2 + y2) { (x2)2 – x2 y2 + (y2)2}]

= x (x2 + y2) (x4 – x2 y2 + y2)

 

(xiii) 2a4 – 128a

= 2a (a3 – 64)

= 2a (a3 – 43)

= 2a (a – 4) (a2 + 4a + 42)

= 2a (a – 4) (a2 + 4a + 16)


  1. Factorise

(i) (1 – a)3 + (3a)3

Solution:

Using x3 + y3 = (x + y) (x2 – xy + y2)

(1 – a)3 + (3a)3

= (1 – a + 3a) [(1 – a)2 – (1 – a) 3a + (3a)2]

= (1 + 2a) [(1 – a)2 – 3a + 3a2 + 9a2]

= (1 + 2a) (1 + a2 – 2a – 3a + 12a2)

= (1 + 2a) (13a2 – 5a + 1)

 

(ii) 8x3 – 27y3

Solution:

= (2x)3 – ( 3y)3

= ( 2x – 3y) [(2x)2 + 2x .3y + (3y)2]

= (2x – 3y) (4x2 + 6xy + 9y2)

 

(iii) z4 x3 + 8y3 z4

Solution:

= z4 (x3 + 8y3)

= z4 [x3 + (2y)3]

= z4 (x + 2y) [x2 – x.2y + (2y)2]

= z4 (x + 2y) [x2 – 2xy + 4y2]

 

(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 ) (x2 + y+ 2xy – (x + y)( xy/3) + x2y2/9 )

 

(v) x6 + y6

Solution:

= (x2)3 – (y2)3

= (x2 – y2) [(x2)3 + x2 y2– (y2)3]

= (x2 – y2) (x4 + x2 y2 + y4)

= (x + y) (x – y) (x2 + x y + y2) (x2 – xy + y2)

 

(vi) a3 – 2√𝟐 b3

Solution:

= a3 – (√2 b)3

= (a –√2b) [a2 + a.√2 b + √2 b2]

= (a –√𝟐 b) (a2 + √𝟐 ab + √𝟐b2)


  1. Factorize the following

(i) x6 – 26x3 – 27

Solution:

put x3 = a

a3 – 26a – 27

= a3 – 27a + a – 27

= a ( a – 27) + 1 (a – 27)

= (a – 27) ( a + 1)

= (x3 – 27) (x3 + 1)

= (x – 3) (x2 + 3x + 9) (x – 1) (x2 – x + 1)

 

(ii) z6 – 63z3 – 64

Solution:

put z3 = a

a3 – 63a – 64

= a3 – 64a + a – 64

= a (a – 64) + 1 (a – 64)

= (a – 64) (a + 1)

= (z3 – 64) (z3 + 1)

= (z – 4) (z2 + 4z + 16) (z + 1) (z2 – z + 1)

 

(iii) a3 – b3 – a + b

Solution:

= (a – b) (a2 + ab + b2) – (a – b)

= (a – b) [ a2 + ab + b2 – 1]

 

(iv) x6 + 7x3 – 8

Solution:

put x3 = a

a3 + 7a – 8

= a3 + 8a – a – 8

= a (a + 8) – 1 (a + 8)

= (a – 1) (a + 8)

= (x3 – 1) (x3 + 8)

= (x – 1) (x2 + x – 1) (x + 2) (x2 – 2x + 4)

 

(v) a3 𝟏/𝐚𝟑 – 2a + 𝟐𝐚

Solution:

= (a – 𝟏/𝐚) (a2 + a. 𝟏/𝐚+ 𝟏/𝐚2 ) – 2 (a – 𝟏/𝐚 )

= (a – 𝟏/𝐚 ) [a2 + 1 + 𝟏/𝐚2 – 2]

= (a – 𝟏/𝐚 ) (a2 + 𝟏/𝐚𝟐 – 1 )


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

  1. a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

We have,

a3 + b3 + c3 – 3abc = (a3 + b3) + c3 – 3abc

[use a3 + b3 = (a + b)3 – 3ab(a + b)]

= (a + b)3 + c3 – 3ab(a + b + c)

[use x3 + y3 = (x + y)(x2 – xy + y2)]

= (a + b + c)[(a + b)2 – (a + b)c + c2] – 3ab(a + b + c)

= (a + b + c)[a2 + b2 + 2ab – ac – bc + c2 – 3ab]

=(a + b + c)(a2 + b2 + c2 –  ab – bc – ca)

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 –  ab – bc – ca)


Corollary: If a + b + c = 0 then a3 + b3 + c3 = 3abc

Proof :

We use the identity

a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 –  ab – bc – ca)

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

a3 + b3 + c3 – 3abc = 0

Hence, a3 + b3 + c3 = 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 = a3 + b3 + 3ab(a+b) = a3 + b3 + 3ab(-c)

This simplifies to

a3 + b3 + c3 – 3abc = 0

or

a3 + b3 + c3 = 3abc


Factorization[class 9] – Exercise 3.2.6

  1. Factorise

(i) a3 – b3 – c3 – 3abc

Solution:

= a3 + (– b3)+(– c3) – 3a(–b) ( –c)

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = a y = –b c = –c

∴ a3 + (– b3)+(– c3) – 3a(–b) (c)

= [a + (–b) +(c)] [a2 + (–b)2 + (–c) – a(–b) (–b)(–c) – (–c)a]

= (a – b – c) (a2 + b2 + c2 + ab – bc + ca)

 

(ii) a3 – b3 + 8c3 + 6abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = a y = –b c = 2c

a3 + (–b)3 + (2c)3 – 3a (–b) (2c)

= [a + (– b) + 2c] [a2 + (-b)2 + (2c)2 – a(– b) – (–b) (2c) – 2c.a]

= (a – b + 2c) (a2 + b2 + 4c2 + ab + 2bc – ca)

 

(iii) 125a3 + b3 + 64c3 – 60abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = 5a y = b c = 4c

(5a)3 + b3 + (4c)3 – 3(5a) b (4c)

= (5a + b + 4c) [(5a)2 + b2 + 16c2 – 5a(b) – b(4c) – 4(c) (5a)]

= (5a + b + 4c) (25a2 + b2+ 16c2 – 5ab – 4bc – 20ca)

 

(iv) 1 + b3 + 8c3 – 6bc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = 1 y = b c = 2c

13 + b3 + (2c)3 – 3 . 1 . b . 2c

= (1 + b + 2c) [12 + b2 + (2c)2 – 1 .b – b.2c – 2c.1]

= (1 + b + 2c) (1 + b2 + 4c2 – b – 2bc – 2c)

 

(v) 8a3 + 125b3 – 64c3 + 120abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – 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) (4a2 + 25b2 + 16c2 – 1ab + 20bc + 8ca)

 

(vi) 2√𝟐 a3 + 16√𝟐 b3 + c3 – 12abc

Solution:

Using x3 + y3 + z3 – 3xyz = (x + y +z) (x2 + y2 + z2 – xy – yz – xz) we get

x = √2a ;y =2√2 b;  c = c

(√2 a)3 + (2 √2 b)3 + c3 – 3(√2 a) (2√2 b)c

= (√2 a + 2√2 b + c) [(√2 a)2 + (2√2 b)2 + c2 -√2 a. 2√2 b –

2√2 b . c – c. √2 a]

= (√𝟐 a + 2√𝟐 b + c) (2a2 + 8b2 + c2 – 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 a3 + b3 + c3 = 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) p3 (q – r)3 + q3 (r – p)3 + (p – q)3

[Hint: apply corollary]

Solution:

Using identify if a + b + c = 0 then a3 + b3 + c3 = 3abc 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

∴ p3 (q – r)3 + q3 (r – p)3 + (p – q)3

= 3p (q – r) q(r – p) r(p – q)

= 3pqr (q – r) (r – p) (p – q)


  1. Find the product using appropriate identity

(i) (a – b – c) (a2 + b2 + c2 + ab – bc – ca)

Solution:

Using (x + y + z) (x2 + y2 + z2 – xy – yz – zx) = x3 + y3 + x3

We get x = a y = –b z = –c

(a – b – c) (a2 + b2 + c2 + ab – bc – ca)

= a3 + (-b)3 + (-c)3

= a3 – b3 – c3 – 3abc

 

(ii) (x – 2y – z) (x2 + 4y2 + z2 + 2xy – 2yz – zx)

Solution:

Using (a + b + c) (a2 + b2 + c2 – ab – bc ca) = a3 + b3 + c3 – 3abc

We get a = x b = –2y c = –z

[x + (–2y) + (–z)] [ x2 +(–2y)2 +(-z)2 – x (-2y) – (-2y) (-z) + (-z)(x)]

= x3 + (–2y)3 + (-z)3 – 3x (–2y) (–8)

= x3 – 8y3 – z3 – 6xyz

 

(iii) (x + y – z) (x2 + y2 + z2 – xy + yz + zx)

Solution:

Using (a + b + c) (a2 + b2 + c2 – ab – bc ca) = a3 + b3 + c3 – 3abc

We get a = x b = y c = –z

[x + y + (–z)] [x2 + y2 + (–z)2 – xy + y(–z) + (–z) x]

= x3 + y3 + (-z)3 – 3(x) (y) (-z)

= x3 + y3 – z3 + 3xyz


  1. Get the factorization

(a + b + c)3 – a3 – b3 – c3 = 3 (a+ b) (b + c) (c + a) writing the expression

(a + b + c)3 – a3 – b3 – c3 = {(a + b + c)3 – a3} – {b3 + c3}

Solution:

We have to prove {(a + b + c)3 – a3} – {b3 + c3}

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

L.H.S = {(a + b + c)3 – a3} – {b3 + c3}

Using identity a3 – b3 = (a – b) (a2 + ab + b2)

= (a + b + c – a) {(a + b + c)2 + (a + b + c)a + a2} – (b – c) (b2 – bc + c2)

= (b + c) [(a2 + b2 + c2 + 2ab + 2bc + 2ca + a2 + ab + ac + a2) – (b2 – bc + c2)

= (b + c) [(3a2 + b2 + c2 + 3ab + 2bc + 3ca] – (b + c) (b2 – bc + c2)

= [3a2 + b2 + c2 + 3ab + 2bc + 3ca – b2 + bc – c2]

= (b + c) (3a2 + 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


  1. If x + y + 4 = 0, find the value of x3 + y3 – 12xy + 64.

Solution:

x + y = – 4

Cubing on both sides

(x + y)3 = (– 4)3

x3 + y3 + 3xy (x + y) = (– 4)3

x3 + y3 + 3xy (– 4)3 = – 64

x3 + y3 – 12xy + 64 = 0


  1. 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

= x3 – (2y)3 – 3x (2y) (x – 2y) = 216

= x3 – 8y3 – 6xy (6) – 216 = 0

= x3 – 8y3 – 36xy – 216 = 0


  1. Without actually calculating the cubes, find the values of the following:

(i) (–12)3 + 73 + 53

Solution:

We have – 12 + 7 + 5 = 0

By the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–12)3 + 73 + 53 = 3 (–12) (7) (5)

= -1260

 

(ii) (28)3 + (–315)3 + (–13)3

Solution:

We find that 28 – 15 – 13 = 0

By the identify if a + b + c = 0 then a3 + b3 + c33abc we get

(28)3 + (–15)3 + (–13)3 = 3 x 28 x (–15) (–13)

= 16380

 

(iii) (–15)3 + 73 + 83

Solution:

Since – 15 + 7 + 8 = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–15)3 + 73 + 83 = 3 (–15) 7 x 8

= – 2520

 

(iv) (–10)3 + 33 + 73

Solution:

Since – 10 + 3 + 7 = 0

Using the identify if a + b + c = 0 then a3 + b3 + c3 – 3abc we get

(–10)3 + 33 + 73 = 3(-10) (3) (7)

= – 630


  1. Factorise the following using the identity

a3 + b3 + c3 – 3abc = 𝟏/𝟐 (a + b + c) [(a – b)2 + (b – c)2 + (c – a)2]

(i) a3 + 8b3 + c3 – 6abc

Solution:

= 1/2 (a + 2b + c) [(a – 2b)2 + (2b – c)2 + (c – a)2]

= 1/2 (a + 2b + c) (a2 + 4b2 – 4ab + 4b2 + c2 – 4bc + c2 +

a2 – 2ac)

= 1/2 (a + 2b + c) (2a2 + 8b2 + 2c2 – 4ab – 4bc – 4ca)

 

(ii) 125a3 + b3 + 64c3 – 60abc

Solution:

= 1/2  (5a + b + 4c) [(5a – b)2 + (b – 4c)2 + (4c – 5a)2]

= 1/2  (5a + b + 4c) (25a2 + b2 – 10ab + b2 + 16c2 – 8bc + 16c2

+ 25a2 – 40ac)

= 1/2 (5a + b + 4c) (50a2 + 2b2 + 32c2 – 10ab – 8bc – 40ac)

 

(iii) 13 + b3 + 8c3 – 6bc

Solution:

= 1/2 (1 + b + 2c) [(1 – b)2 + (b – 2c)2 + (2c – 1)2]

= 1/2 (1 + b +2c) (1 + b2 – 2b + b2 – 4bc + 4c2 + 4c2 + 1 – 4c)

= 1/2 (1 + b + 2c) (2 + 2b2 + 8c2 – 2b – 4bc – 4c)

 

(iv) 125 – 8x3 – 27y3 – 90xy

Solution:

= 1/2 (5 – 2x – 3y) [(5 – 2x)2 + (–2x – 3y)2 + (–3y – 5)2]

= 1/2 (5 – 2x – 3y) (25 + 4x2 – 20x + 4x2 + 9y2 – 12xy + 9y2 +

25 – 30y)

= 1/2 (5 –2x – 3y) (50 + 8x2 + 18y2 – 20x – 12xy – 30y)


  1. 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 a3 + b3 + c3 – 3abc we get

(x – 2y)3 + (2y – 3z)3 + (3z – x)3

= 3(x – 2y) (2y – 3z) (3z – x)

 

(ii) [(𝐱𝟐− 𝐲𝟐)𝟑+ (𝐲𝟐− 𝐳𝟐)𝟑+(𝐳𝟐− 𝐱𝟐)𝟑(𝐱 − 𝐲)𝟑+ (𝐲 − 𝐳)𝟑+(𝐳 − 𝐱)𝟑]/[(x – y)3 (y – z)3 (z – x)3]

Solution:

We find that x2 – y2 + y2 – z2 + z2 – y2 = 0

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

a + b + c = 0 then a3 + b3 + c33abc we get

[(x2− y2)3+ (y2− z2)3+(z2− x2)3] /[(x − y)3+ (y − z)3+(z − x)3]

= [3(x2− y2)+ (y2− z2)+(z2− x2)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)]

[∵ a2 – b2 = (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 a3 + b3 + c3 – 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 a3 + b3 + c3 – 3abc we get

(a – 3b)3 + (3b – c)3 + (c – a)3

= 3 (a – 3b) (3b – c) (c – a)


  1. Factorise the following:

(i) (x – y – z)3 – x3 + y3 + z3

Solution:

= [x + (–y) + (–z)]3 – x3 – (–y)3 – (–z)3

Using the identify if (a + b + c)3 – a3 – b3 – c3 = 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 – a3 + b3 + 1

Solution:

= [a + (–b)3 + (–1)3 – a3 – (–b)3 – (–1)3

Using the identify if (x + y + z)3 – x3 – y3 – z3 = 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 – 8x3 – y3 + z3

Solution:

= [2x + y + (–z)]3 – (2x)3 – y3 – (–z)3

Using the identify if (a + b + c)3 – a3 – b3 – c3 = 3(a + b) (b +c) (c +a) we get

a = 2x b = y c = – z

= 3(2x + y) 2017 [(–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 – a3 𝟏/𝟖 b3 𝟏/𝟐𝟕 c3

Solution:

= (a – (𝟏/𝟐 b) + 𝟐/𝟑 c )3 – a3 – ( 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/3b/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 – a3 – b3 – c3 = 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 – a3 – b3 – c3 = 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)


  1. Find the factors of the following numbers

(i) 303 – 123 – 103 – 83

Solution:

We find that 12 + 10 + 8 = 30

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 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

= 24 x 33. 5.11

 

(ii) 853 – 683 + 53 – 223

Solution:

We find that 68 – 5 + 22 = 85

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 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

= 2.35.5.7.17

 

(iii) 1003 – 493 + 103 – 613

Solution:

We find that 49 – 10 + 61 = 100

Using the identify

(a + b + c)3 – a3 – b3 – c3 = 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

= 2.33.5.11.13.17

 

  1. 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 {y2 + z2 + x2 + 2yz – 2zx – 2xy – (yz + z2 – zx + xy + zx – x2 – y2

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

= [2z (3x2 + 3y2 + z2 + 6xy) – 2z (y2 + z2 + x2 + 2yz – 2zx – 2xy – yz – z2

+ zx – xy – zx + x2 + y2 + yz – xy + z2 + x2 + y2 + 2xy – 2xy – 2yz)

= (x2z + 6y2z + 2z3 + (x + y) – 2z (3x2 + 3y2 – 6xy)

= 6x2z + 6y2z + 12xyz – 6x2z – 6y2z – 2z3 + 12xyz

= 24xyz

= R.H.S