Q. 34

# Factorise:

(i) 1 + 64π₯^{3}

(ii) π^{3}β2β2π^{3}

Answer :

(i) Given that: 1 + 64x^{3}

= 1 + (4x)^{3}

= (1 + 4x)[(1)^{2} β (1)(4x) + (4x)^{2}]

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

= (1 + 4x)(1 β 4x + 16x^{2})

(ii) Given that: a^{3} β 2β2b^{3}

= (a)^{3} β (β2b)^{3}

= (a β β2b) [(a)^{2} + (a)(β2b) + (β2b)^{2}]

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

= (a β β2b)(a^{2} + β2ab + 2b^{2})

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