Question:
Factorise:
$2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c$
Solution:
$2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c$
$=(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3(\sqrt{2} a)(\sqrt{3} b) c$
We know
$x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$
$x=\sqrt{2} a, y=\sqrt{3} b, z=c$
$(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3(\sqrt{2} a)(\sqrt{3} b) c$
$=(\sqrt{2} a+\sqrt{3} b+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6 a b}-\sqrt{3 b c}-\sqrt{2} a c\right)$
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