Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ Find coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$
how to do it? I think it should be $3^6$ since $(3x^2)^6=3^6x^8$. (this is false)
Is this true?

Let $S:=(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)$.  Then $$\begin{align}S(1+x)&=1-x+x^2-x^3+x^4-x^5+x^6+7x^7\\
&=\frac{1+x^7}{1+x}+7x^7\end{align}$$
So $$\begin{align} S&=\frac{1+x^7}{(1+x)^2}+\frac{7x^7}{1+x} \end{align}$$
Then we have $$ S^6=\sum_{r=0}^6{n\choose r}\frac{(1+x^7)^r(7x^7)^{n-r}}{(1+x)^{2r+n-r}} $$
The observant eye will note that $n-r\le 1$, otherwise $\deg(7x^7)^{n-r}>8$, so we look at: 
$$ 6\frac{(1+x^7)^5\cdot 7x^7}{(1+x)^{11}}+\frac{(1+x^7)^6}{(1+x)^{12}} $$
One may note that $\frac{1}{(1-x)^n}=\sum_{k=0}^\infty {k+n-1\choose n-1}x^k$, so in the first term we have $$42x^7(1+x^7)^5(1-11x+\ldots)\to-462x^8$$
And in the second term: $$ (1+6x^7+\ldots)(1-12x+\ldots+{19\choose 11}x^8-\ldots )\to -72x^8+75582x^8 $$
So $75582-462-72=75048$ is the coefficient of $x^8$.
