Factorize each of the following expressions:$ 2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c $

Given:

\( 2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c \)

To do:

We have to multiply the given expressions.

Solution:

We know that,

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

$a^3 + b^3 + c^3 = 3abc$ if $a + b + c = 0$

Therefore,

$2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c=(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+(c)^{3}-3 \times \sqrt{2} a \times \sqrt{3} b \times c$

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

$=(\sqrt{2} a+\sqrt{3} b+c)(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} c a)$

Hence, $2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c=(\sqrt{2} a+\sqrt{3} b+c)(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} c a)$.