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 $5.$

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 
