If $(1-x^3)^n=\sum_{r=0}^{n} a_r x^r (1-x)^{3n-2r}$, then prove that $a_r=\binom{n}{r} 3^r$. If $$(1-x^3)^n=\sum_{r=0}^{n} a_r x^r (1-x)^{3n-2r}$$ then prove that  $$a_r=\binom{n}{r} 3^r.$$
I have factored $1-x^3=(1-x)(1+x+x^2)$ and cancelled out one factor $(1-x)^n$ from both sides after which we get
$$(1+x+x^2)^n=\sum_{r=0}^{n} a_r x^r (1-x)^{2n-2r}.$$ 
Any clue here?

I begin with the answer :
$\sum_0^n \binom{n}{r} 3^r x^r (1-x)^{3n-2r} =
 \sum_0^n \binom{n}{r} (3x)^r ((1-x)^2)^{n-r} (1-x)^n$
which is $(1-x)^n ((3x) + (1-x)^2)^n = (1-x)^n (1+x+x^2)^n$
I tried to make something like $\binom{n}{r} A^r B^{n-r}$
