Q. 34

Factorise:

(i) 1 + 64π‘₯3

(ii) π‘Ž3βˆ’2√2𝑏3

Answer :

(i) Given that: 1 + 64x3


= 1 + (4x)3


= (1 + 4x)[(1)2 βˆ’ (1)(4x) + (4x)2]


[∡ a3 + b3 = (a + b)(a2 – ab + b2) ]


= (1 + 4x)(1 – 4x + 16x2)


(ii) Given that: a3 βˆ’ 2√2b3


= (a)3 βˆ’ (√2b)3


= (a βˆ’ √2b) [(a)2 + (a)(√2b) + (√2b)2]


[∡ a3 – b3 = (a + b)(a2 + ab + b2) ]


= (a βˆ’ √2b)(a2 + √2ab + 2b2)


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