Question

Factorize: $2\sqrt{2}{a}^3+8b^3-27c^3+18\sqrt{2}abc.$

• A. $(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2+2\sqrt{2}ab+6bc+3\sqrt{2}ac)$
• B. $(\sqrt{2}a+2b-3c)(a^2+4b^2+9c^2+2\sqrt{2}ab+6bc+3\sqrt{2}ac)$
• C. $(\sqrt{2}a+2b-3c)(a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$
• D. $(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$

Answer 1

The correct option is D $(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$

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