Given ,
$\large (\frac{1-x^{6}}{1-x})^{3}$
=$\large (1-x^{6})^{3}(1-x)^{-3}$
=$\large (1-3x^{6}+3x^{12}-x^{18})(1-x)^{-3}$
Now we need to expand $\large (1-x)^{-3}$.
$\large (1-x)^{-3}$ [By extended binomial theorem]
=$\large (1+\binom{-3}{1}(-x)+\binom{-3}{2}x^{2}+\binom{-3}{3}(-x)^{3}+.......)$
Now, $\large \binom{-n}{r}=(-1)^{r}\binom{n+r-1}{r}$ [only if n is integer ]
=$\large (1+3x+6x^{2}+10x^{3}+15x^{4}+21x^{5}+28x^{6}+36x^{7}+45x^{8}+........)$
So putting this value in the expression ,
=$\large (1-3x^{6}+3x^{12}-x^{18})$ . $\large (1+3x+6x^{2}+10x^{3}+15x^{4}+21x^{5}+28x^{6}+36x^{7}+45x^{8}+........)$
So Coefficient of $\large x^{8}$ is $(45+(-3)*6)=27$ .