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 $5.$
 A: You're almost there.  You've shown that the coefficient on $x^5$ is $0.$
A: there is no $x^5$ th term in the expanding of the binom.Because $${ \left( 4{ x }^{ 2 }-\frac { 1 }{ { x }^{ 3 } }  \right)  }^{ 8 }=...A{ x }^{ 5 }+..\\  \binom {8} {k} { \left( 4{ x }^{ 2 } \right)  }^{ 8-k }{ \left( -{ x }^{ -3 } \right)  }^{ k }=A{ x }^{ 5 }\\ { x }^{ 16-2k-3k }={ x }^{ 5 }\\ 16-5k=5\\ 5k=11\\ k=2,2\\ $$ but $k$ should be positive  integer number 
A: 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:
$$
\begin{array}{ccr}
 p & q & 2p-3q \\\hline
 0 & 8 & -24 \\
 1 & 7 & -19 \\
 2 & 6 & -14 \\
 3 & 5 & -9 \\
 4 & 4 & -4 \\
 5 & 3 & 1 \\
 6 & 2 & 6 \\
 7 & 1 & 11 \\
 8 & 0 & 16 \\
\end{array}
$$
A: The term in given expansion containing $x^5$ $=^8C_r(4x^2)^r(-1/x^3)^{8-r} = ^8C_r4^r(-1)^{8-r}x^{2r-3(8-r)} = ^8C_r4^r(-1)^{8-r}x^{5r-24}$
For $x^5$, $5r-24=5$, so $r=29/5$ which is not a whole number. So there is no term in the expansion of $(4x^2-(1/x^3))^8$ having term with $x^5$. So coefficient of $x^5$ in the given expansion is $\color{red}{0}$
