$a^3+8 b^3+64 c^3-24 a b c$
$=a^3+(2 b)^3+(4 c)^3-3 \times a \times 2 b \times 4 c$
$=(a+2 b+4 c)\left[a^2+(2 b)^2+(4 c)^2-a \times 2 b-2 b \times 4 c-4 c \times a\right]$
$=(a+2 b+4 c)\left(a^2+4 b^2+16 c^2-2 a b-8 b c-4 c a\right)$
We know that:
Since $a^3-b^3=(a-b)\left(a^2+a \times b+b^2\right)$
Let us rewrite
$8\text{x}^3-\frac{1}{27\text{y}^3}$
$=(2\text{x})^3-\Big(\frac{1}{3\text{y}}\Big)^3$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\bigg[(2\text{x})^2+2\text{x}\times\frac{1}{3\text{y}}+\Big(\frac{1}{3\text{y}}\Big)^2\bigg]$
$=\Big(2\text{x}-\frac{1}{3\text{y}}\Big)\Big(4\text{x}^2+\frac{2\text{x}}{3\text{y}}+\frac{1}{9\text{y}^2}\Big)$