Prove the following



(i) 1 + 64x3          

(ii) a3 -2√2b3

Thinking Process

(i) Firstly, adjust the given number either in the farm 0f a3 + b3 or a3 -b3

(ii) Further, use any of the identities i.e., a3 + b3 =(a + b)(a2+b2-ab) and a3 -b3 =(a-b)(a2 + b2 + ab), then simplify it, to get the factor.


(i) $1+64 x^{3}=(1)^{3}+(4 x)^{3}$

$=(1+4 x)\left[(1)^{2}-(1)(4 x)+(4 x)^{2}\right]$

[using identity, $\left.a^{3}-b^{3}=(a+b)\left(a^{2}-a b+b^{2}\right)\right]$

$=(1+4 x)\left(1-4 x+16 x^{2}\right)$

(ii) $a^{3}-2 \sqrt{2} b^{3}=(a)^{3}-(\sqrt{2} b)^{3}$

$=(a-\sqrt{2} b)\left[a^{2}+a(\sqrt{2} b)+(\sqrt{2} b)^{2}\right]$

[using identity, $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)\right]$

$=(a-\sqrt{2} b)\left(a^{2}+\sqrt{2} a b+2 b^{2}\right)$

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