Determine the co-efficient of $x^{97}$ in the expansion of $(1-x)^{135}(1+x+x^2)^{135}$.

Question

Determine the co-efficient of $x^{97}$ in the expansion of $(1-x)^{135}(1+x+x^2)^{135}$.

I know to use the binomial theorem, but I am having a difficult time simplifying.

Any help is appreciated, thanks!

Answer 1

You get $(1-x^3)^{135}$. Since $3\not\mid97$, the coefficient of $x^{97}$ is $0$.

Answer 2

The problem becomes rather simple when you realize that$$(1-x)(1+x+x^2)=1-x^3$$

Using the binomial expansion, it's easy to see that$$\begin{align*}(1-x^3)^{135} & =\sum\limits_{r=0}^{135}\binom {135}r(-1)^rx^{3r}\end{align*}$$The coefficient $x^{97}$ is when $r=\tfrac {97}3$. But since $r$ isn't an integer, then that means the coefficient is just simply zero because each multiple advances by a factor of three.