$[x^8] (1- x^6)^3 (1 – x)^{-3}$

$[x^8] \sum_{r=0}^{3} \binom{3}{r}(-1)^r x^{6r} \sum_{k=0}^{\infty} \binom{3+k-1}{k} x^{k}$

So, $r=0,k=8$ and $r=1,k=2$

Hence, Answer is $\binom{3}{0}\binom{3+8-1}{8} - \binom{3}{1}\binom{3+2-1}{2} = 27$

$[x^8] \sum_{r=0}^{3} \binom{3}{r}(-1)^r x^{6r} \sum_{k=0}^{\infty} \binom{3+k-1}{k} x^{k}$

So, $r=0,k=8$ and $r=1,k=2$

Hence, Answer is $\binom{3}{0}\binom{3+8-1}{8} - \binom{3}{1}\binom{3+2-1}{2} = 27$