<s>[INST] Finding the coefficient of $x^{10}$ in $(1+x^2-x^3)^8$ 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.
 [/INST] The pertinent part of the polynomial expansion can be done by hand without too much fuss:
$$\begin{align}
(1+x^2-x^3)^8&=(1+x^2)^8-8x^3(1+x^2)^7+{8\choose2}x^6(1+x^2)^6-{8\choose3}x^9(1+x^2)^5+\cdots\\
&=(1+x^2)^5((1+x^2)^3-8x^3(1+x^2)^2+28x^6(1+x^2)-56x^9)+\cdots\\
&=(1+x^2)^5((1+3x^2+3x^4+x^6)-8x^3(1+2x^2+x^4)+28x^6(1+x^2)-56x^9)+\cdots\\
&=(1+5x^2+10x^4+10x^6+5x^8+x^{10})(1+3x^2-8x^3+3x^4-16x^5+29x^6-8x^7+28x^8-56x^9)+\cdots\\
&=\cdots+(5\cdot28+10\cdot29+10\cdot3+5\cdot3+1\cdot1)x^{10}+\cdots\\
&=\cdots+(140+290+30+15+1)x^{10}+\cdots\\
&=\cdots+476x^{10}+\cdots
\end{align}$$
</s>