Question:
Find the value
$2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6 a b c}$
Solution:
$=(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3 \times \sqrt{2 a} \times \sqrt{3 b} \times c$
$=(\sqrt{2 a}+\sqrt{3 b}+c)\left((\sqrt{2 a})^{2}+(\sqrt{3 b})^{2}+c^{2}-(\sqrt{2 a})(\sqrt{3 b})-(\sqrt{3 b}) c-(\sqrt{2 a} c)\right)$
$=(\sqrt{2 a}+\sqrt{3 b}+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} a c\right)$
$\therefore 2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6 a b c}$
$=(\sqrt{2 a}+\sqrt{3 b}+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6} a b-\sqrt{3} b c-\sqrt{2} a c\right)$