<s>[INST] Coefficient of $x^5$ in $(4x^2- \frac{1}{x^3})^8$ I took one $x$ out of bracket, and I got $\frac{1}{x^{24}}  (4x^5-1)^8.$ Now, we get $x$ to the power of if we get $x$ to the power $29,$ then we divide it by $24,$ we get $x$ to the power $5,$ but i can't get $29,$ by multiplying $5.$
 [/INST] There are no odd powers in this expansion.
The generic term will look like
$$
 a_{p,q} \left( 4x^{2} \right)^{p} \left( \frac{1}{x^{3}} \right)^{q}
$$
where $p+q=8$.
The powers in the generic term reduce to
$$
x^{2p-3q} = x^{2p-3(8-p)} = x^{5p-24}
$$
The question is this: For what integer is this equation valid:
$$
 5p = 24?
$$
There is no such integer. Therefore, there is no term like $x^{5}$.

A brute force method computes all the powers:
$$
\begin{array}{ccr}
 p & q & 2p-3q \\\hline
 0 & 8 & -24 \\
 1 & 7 & -19 \\
 2 & 6 & -14 \\
 3 & 5 & -9 \\
 4 & 4 & -4 \\
 5 & 3 & 1 \\
 6 & 2 & 6 \\
 7 & 1 & 11 \\
 8 & 0 & 16 \\
\end{array}
$$
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