We know that $\frac{1}{(1-x)^n} = \sum_{k=0}^\infty \binom{k+n-1}{n-1} x^k$
I also worked out $(1-x^3)^n$ using the binomial theorem and got $(1-x^3)^n = \sum_{i=0}^\infty (-1)^i \binom{n}{n-i} x^{3i}$
I'm not sure what to do with these to get $a_k$ from $\sum_{k=0}^\infty a_k x^k$ or if these are even what I need to solve the problem
Any help is appreciated
Hint: \begin{align*} \left(\frac{1-x^3}{1-x}\right)^n=\left(\frac{(1-x)\left(1+x+x^2\right)}{1-x}\right)^n=\left(1+x+x^2\right)^n \end{align*}