## Book: NCERT - Mathematics

### Chapter: 14. Factorisation

#### Subject: Maths - Class 8th

##### Q. No. 4 of Exercises 14.2

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##### Factorise.(i) (ii) (iii) (iv) (v)

(i) a⁴ - b⁴ = (a2)2 - (b2)2

Using identity a2 - b2 = (a + b)(a - b)

Here a = a2 ; b = b2

(a2)2 - (b2)2 = (a2 + b2) (a2 - b2)

Again Using identity a2 - b2 = (a + b)(a - b)

Here a = a ; b = b

a2 - b2 = (a + b)(a - b)

(a2)2 - (b2)2 = (a2 + b2) (a + b)(a - b)

(ii) p⁴ - 81⁴ = (p2)2 - (32)2

Using identity a2 - b2 = (a + b)(a - b)

Here a = p2 ; b = 92

(p2)2 - (92)2 = (p2 + 92) (p2 - 92)

Again Using identity a2 - b2 = (a + b)(a - b)

Here a = p ; b = 3

p2 - 32 = (p + 3)(p - 3)

(p2)2 - (92)2 = (p2 + 92) (p + 3)(p - 3)

(iii) x⁴ - (y + z)⁴ = (x2)2 - {(y + z)2}2

Using identity a2 - b2 = (a + b)(a - b)

Here a = x2 ; b = (y + z)2

(x2)2 - (y + z2)2 = {x2 + (y + z)2} {x2 - (y + z)2}

Again Using identity a2 - b2 = (a + b)(a - b)

Here a = x ; b = y + z

x2 - (y + z)2 = {x + (y + z)}{(x – (y + z)}

(x2)2 - (y + z2)2 = {x2 + (y + z)2} (x + y + z)(x – y - z)

(iv) x⁴ - (x - z)⁴ = (x2)2 - {x - z)2}2

Using identity a2 - b2 = (a + b)(a - b)

Here a = x2 ; b = (x - z)2

(x2)2 - (x - z2)2 = {x2 + (x - z)2} {x2 - (x - z)2}

Again Using identity a2 - b2 = (a + b)(a - b)

Here a = x ; b = x - z

x2 - (x - z)2 = {x + (x - z)}{(x – (x - z)}

(x2)2 - (x - z2)2 = {x2 + (x - z)2} (x + x - z)(x – x + z)

(x2)2 - (x - z2)2 = {x2 + (x - z)2} (2x - z)(z)

(x2)2 - (x - z2)2 = (2x2 -2xz + z2) (2x - z)(z) [using (a + b)2 = a2 + b2 + 2ab]

(v) a⁴ - 2a2b2 + b⁴ = (a2)2 -2 × a× b + (b2)2

(a2)2 -2 × a× b + (b2)2 = (a2 - b2)2 [using (a - b)2 = a2 -2a2b2 + b2]

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