Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$

Question

The coefficient of $x^{10}$ in $(1+x^2-x^3)^8$. I tried to factor it into two binomials but it became way to long to solve by hand.

Answer 1

More generally, the coefficient of $x^{n}$ is given by $$[x^{n}](1+x^2(1-x))^8=[x^{n}]\sum_{k=0}^8\binom{8}{k}x^{2k}(1-x)^k =(-1)^n\sum_{n/3\leq k\leq \min(n/2,8)}\binom{8}{k}\binom{k}{n-2k}.$$ For $n=10$ we get $$\sum_{4\leq k\leq 5}\binom{8}{k}\binom{k}{10-2k}=\binom{8}{4}\binom{4}{2}+\binom{8}{5}\binom{5}{0}=476.$$

Answer 2

$10=2\cdot 5$ or $2\cdot 3+2\cdot 2$. Hence $$ \binom{8}{5}+\binom{8}{2}\binom{6}{2}=476. $$

Answer 3

The pertinent part of the polynomial expansion can be done by hand without too much fuss:

$$\begin{align} (1+x^2-x^3)^8&=(1+x^2)^8-8x^3(1+x^2)^7+{8\choose2}x^6(1+x^2)^6-{8\choose3}x^9(1+x^2)^5+\cdots\\ &=(1+x^2)^5((1+x^2)^3-8x^3(1+x^2)^2+28x^6(1+x^2)-56x^9)+\cdots\\ &=(1+x^2)^5((1+3x^2+3x^4+x^6)-8x^3(1+2x^2+x^4)+28x^6(1+x^2)-56x^9)+\cdots\\ &=(1+5x^2+10x^4+10x^6+5x^8+x^{10})(1+3x^2-8x^3+3x^4-16x^5+29x^6-8x^7+28x^8-56x^9)+\cdots\\ &=\cdots+(5\cdot28+10\cdot29+10\cdot3+5\cdot3+1\cdot1)x^{10}+\cdots\\ &=\cdots+(140+290+30+15+1)x^{10}+\cdots\\ &=\cdots+476x^{10}+\cdots \end{align}$$