Question

Factorise – 2 √2a3+8b3-27c3+18 2√abc

Answer 1

this is your answer...

Answer 2

Answer:

$(\sqrt{2}a)^3+(2b)^3+(-3c)^3-3(\sqrt{2}a)(2b)(-3c)=(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$

Step-by-step explanation:

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

We have to factorize the given expression.

Using algebraic identity

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

Comparing left side of identity with given expression

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

Thus, $a=\sqrt{2}a\\\\ b=2b\\\\ c=-3c$

Substitute, we have,

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

On simplify, we get,

$(\sqrt{2}a)^3+(2b)^3+(-3c)^3-3(\sqrt{2}a)(2b)(-3c)=(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$

Thus the factorize form of given expression $2\sqrt{2}a^3+8b^3-27c^3+18\sqrt{2}abc$ is $(\sqrt{2}a+2b-3c)(2a^2+4b^2+9c^2-2\sqrt{2}ab+6bc+3\sqrt{2}ac)$