Question

Question 36
Factorise:
(i) $a^3-8b^3-64c^3-24abc$

(ii) $2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc$

Answer 1

(i) $a^3-8b^3-64c^3-24abc=(a{)}^3+(-2b{)}^3+(-4c{)}^3-3\times (a)\times (-2b)\times (-4c)$
$=(a-2b-4c)\left[\right(a{)}^2+(-2b{)}^2+(-4c{)}^2-a(-2b)-(-2b)(-4c)-(-4c)\left(a\right)]$
$[usingtheidentity,a^3+b^3+c^3-3abc=(a+b+c\left)\right(a^2+b^2+c^2-ab-bc-ca\left)\right]$
$=(a-2b-4c)(a^2+4b^2+16c^2+2ab-8bc+4ac)$

(ii) $2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc$
$=(\sqrt{2}a{)}^3+(2b{)}^3+(-3c{)}^3-3(\sqrt{2}a\left)\right(2b\left)\right(-3c)$
$=(\sqrt{2}a+2b-3c)\left[\right(\sqrt{2}a{)}^2+(2b{)}^2+(-3c{)}^2-\left(\sqrt{2}a\right)\left(2b\right)-\left(2b\right)(-3c)-(-3c)\left(\sqrt{2}a\right)]$
$[usingtheidentity,a^3+b^3+c^3-3abc=(a+b+c\left)\right(a^2+b^2+c^2-ab-bc-ca\left)\right]$
$=(\sqrt{2}a+2b-3c)[2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac]$