Prove the following

(i) $a^{3}-8 b^{3}-64 c^{3}-24 a b c=(a)^{3}+(-2 b)^{3}+(-4 c)^{3}-3 \times(a) \times(-2 b) \times(-4 c)$

$=(a-2 b-4 c)\left[(a)^{2}+(-2 b)^{2}+(-4 c)^{2}-a(-2 b)-(-2 b)(-4 c)-(-4 c)(a)\right]$

[using identity, $\left.a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

$=(a-2 b-4 c)\left(a^{2}+4 b^{2}+16 c^{2}+2 a b-8 b c+4 a c\right)$

(ii) $2 \sqrt{2} a^{3}+8 b^{3}-27 c^{3}+18 \sqrt{2} a b c$

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

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

[using identity, $\left.a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)\right]$

$=(\sqrt{2} a+2 b-3 c)\left[2 a^{2}+4 b^{2}+9 c^{2}-2 \sqrt{2} a b+6 b c+3 \sqrt{2} a c\right]$