Question

Factorise:
$2\sqrt{2}{a}^3+3\sqrt{3}{b}^3+c^3-3\sqrt{6}abc$

Answer 1

$$\begin{gathered} 2\sqrt{2}{a}^3+3\sqrt{3}{b}^3+c^3-3\sqrt{6}abc \\ ={\left(\sqrt{2}a\right)}^3+{\left(\sqrt{3}b\right)}^3+c^3-3\left(\sqrt{2}a\right)\left(\sqrt{3}b\right)c \\ \mathrm{We}\mathrm{know} \\ {x}^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right) \\ x=\sqrt{2}a,y=\sqrt{3}b,z=c \\ {\left(\sqrt{2}a\right)}^3+{\left(\sqrt{3}b\right)}^3+c^3-3\left(\sqrt{2}a\right)\left(\sqrt{3}b\right)c \\ =\left(\sqrt{2}a+\sqrt{3}b+c\right)\left(2a^2+3b^2+c^2-\sqrt{6ab}-\sqrt{3bc}-\sqrt{2}ac\right) \end{gathered}$$