Question

Question 34
Factorise:
(i) $1+64x^3$

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

Answer 1

(i) $1+64x^3=(1{)}^3+(4x{)}^3$
$=(1+4x)\left[\right(1{)}^2–\left(1\right)\left(4x\right)+\left(4x{)}^2\right]$
$[Usingtheidentity,a^3–b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]$
$=(1+4x)(1–4x+16x^2)$

(ii) $a^3-2\sqrt{2}{b}^3=(a{)}^3–(\sqrt{2}b{)}^3$
$=(a-\sqrt{2}b)[a^2+a(\sqrt{2}b)+(\sqrt{2}b{)}^2]$
$[Usingtheidentity,a^3–b^3=(a–b\left)\right(a^2+ab+b^2\left)\right]$
$(a-\sqrt{2}b)(a^2+\sqrt{2}ab+2b^2)$