$$(1-x^3)^{n} = \sum_{r=0}^n a_{r}x^{r}(1-x)^{3n-2r}$$ where n belongs to the set of natural numbers.
Clearly, it should be solved using Binomial theorem. But i don't seem to find out how. Please help me by providing a solution and do mention the steps. Thanks.
HINT:
For $x\ne1,$ $$\dfrac{1-x^3}{(1-x)^3}=\dfrac{1+x+x^2}{(1-x)^2}=\dfrac{(1-x)^2+3x}{(1-x)^2}=1+3\cdot\dfrac x{(1-x)^2}$$
$$\implies\left(\dfrac{1-x^3}{(1-x)^3}\right)^n=\left(1+3\cdot\dfrac x{(1-x)^2}\right)^n$$
Now look at the Binomial series