Q. 34

# Factorise:(i) 1 + 64π₯3(ii) π3β2β2π3

(i) Given that: 1 + 64x3

= 1 + (4x)3

= (1 + 4x)[(1)2 β (1)(4x) + (4x)2]

[β΅ a3 + b3 = (a + b)(a2 β ab + b2) ]

= (1 + 4x)(1 β 4x + 16x2)

(ii) Given that: a3 β 2β2b3

= (a)3 β (β2b)3

= (a β β2b) [(a)2 + (a)(β2b) + (β2b)2]

[β΅ a3 β b3 = (a + b)(a2 + ab + b2) ]

= (a β β2b)(a2 + β2ab + 2b2)

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