<s>[INST] The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
 [/INST] Here is a different way. Let $f(x)=(1+x)^3(2+x^2)^{10}$. Then \begin{align}
f'(x)&=3(1+x)^2(2+x^2)^{10}+20x(1+x)^3(2+x^2)^9\\
f''(x)&=...\\
f'''(x)&=6 \left(x^2+2\right)^{10}+360 x (x+1) \left(x^2+2\right)^9+9 (x+1)^2
   \left(20 \left(x^2+2\right)^9+360 x^2
   \left(x^2+2\right)^8\right)+(x+1)^3 \left(1080 x
   \left(x^2+2\right)^8+5760 x^3 \left(x^2+2\right)^7\right)
\end{align}
Hence the coefficient you are after is $$\frac16f'''(0)=16384$$
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