Formal power series coefficient problem Find the coefficient of:
$[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}$
I have figured out that I need to use this identity:
$(1-x)^{-k} = \sum\limits_{i>=0}  \binom {n+k-1} {k-1} x^n  $
But I have no clue how to proceed with this, I have been stuck with this for hours please help.

$$
\begin{align}
[x^{33}](x+x^3)(1+5x^6)^{-13}(1-8x^9)^{-37}
&=[x^{32}](1+5x^6)^{-13}(1-8x^9)^{-37}\\
&+[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{30}](1+5x^6)^{-13}(1-8x^9)^{-37}\\[6pt]
&=[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37}
\end{align}
$$
The binomial theorem gives
$$
\begin{align}
(1+5x^2)^{-13}
&=\sum_{j=0}^\infty\binom{-13}{j}\left(5x^2\right)^j\\
&=\sum_{j=0}^\infty\binom{j+12}{12}\left(-5x^2\right)^j\\
\end{align}
$$
and
$$
\begin{align}
(1-8x^3)^{-37}
&=\sum_{k=0}^\infty\binom{-37}{k}\left(-8x^3\right)^k\\
&=\sum_{k=0}^\infty\binom{k+36}{36}\left(8x^3\right)^k\\
\end{align}
$$
Therefore,
$$
\begin{align}
[x^{10}](1+5x^2)^{-13}(1-8x^3)^{-37}
&=\sum_{2j+3k=10}\binom{j+12}{12}\binom{k+36}{36}(-5)^j8^k\\
&=\overbrace{\binom{17}{12}\binom{36}{36}(-5)^58^0}^{j=5,\,k=0}+\overbrace{\binom{14}{12}\binom{38}{36}(-5)^28^2}^{j=2,\,k=2}\\[15pt]
&=83019300
\end{align}
$$

Given
Coefficient[Series[(x+x^3)(1+5x^6)^-13(1-8x^9)^-37,{x,0,33}],x,33]
Mathematica returns 83019300
