<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] Let's try it straight from the binomial theorem:
$$
(1+x)^3\cdot(2+x^2)^{10} = \sum_{j=0}^3 {3 \choose j} x^{3-j} \cdot \sum_{k=0}^{10} {10 \choose k} 2^k (x^2)^{10-k}.
$$
Good so far?  Now, note these terms can be arranged as
$$
\sum_{j=0}^3 \sum_{k=0}^{10} {3 \choose j} {10 \choose k}2^k x^{3-j} (x^2)^{10-k}.
$$
Note the $x$ terms combine as 
$$
x^{23 - j - 2k}.
$$
We want the exponent of $x$ to be $3$, which means we want
$$
23 - j - 2k  = 3 \iff j + 2k = 20,
$$
which is satisfied only when $j = 0$ and $k = 10$, and when $j = 2$ and $k = 9$.  Thus the coefficient of $x^3$ is
$$
{3 \choose 0}{10 \choose 10}2^{10} + {3 \choose 2}{10 \choose 9}2^{9} = 16384.
$$
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