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?
 A: 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 $x^8$.
A: Notice 
$$\frac{1}{(1+x)^2} = 1 -2x+3x^2-4x^3+5x^4-6x^5+7x^6-8x^7 + 9x^8 + O(x^9)$$
So the coefficient of $x^8$ in $(1-2x+3x^2-4x^3+5x^4-6x^5+7x^6)^6$ is the same as the
one in
$$\left[\frac{1}{(1+x)^2} + \big(8x^7 - 9x^8\big)\right]^6
= \frac{1}{(1+x)^{12}} + \binom{6}{1}\frac{8x^7-9x^8}{(1+x)^{10}} + O(x^9)
$$
Above equality is true because when we expand LHS by binomial theorem, terms containing factor $(8x^7-9x^8)^k$
is of the order $o(x^{13})$ for $k > 1$. Since 
$$\frac{1}{(1+x)^{\alpha}} = \sum_{n=0}^{\infty} (-1)^n \binom{\alpha+n-1}{n}x^n$$
The coefficient we want is
$$(-1)^8\binom{12+8-1}{8} + 6\times\left[8 \times (-1)^1\binom{10+1-1}{1} - 9\right]\\
= \binom{19}{8} - 6\times 89 =  75582 - 534 = 75048 $$
