# FACTORIZATION EXERCISE 3.2.6 – Class 9

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