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.

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