The cofficient of $x^{32}$in the expansion of $( x^4 - \frac {1}{x^3})^{15}$is

$\textbf{Step 1: Find value of general term for expansion}$

$\text{Given equation is}$

$\bigg ( x^4 -\dfrac {1}{x^3}\bigg)^{15}$

$\text{General term of expansion $ (p+q)^n$ is given by-}$

$T_n= T_{r +1 }=^nC_r p^{ (n-r)}.q^r$

$\text{Comparing given equation with standard form, p=x$^4$and q=-1/x$^3$ and n=15}$

$\Rightarrow T_n=T_{r+1}$

$= ^{15}C_r (x^4)^{(15-r)} .\bigg (\dfrac {-1}{x^3}\bigg)^r$

$= (-1)^r. ^{15}C_r. x^{(60-4r)} .\bigg (\dfrac {1}{x^3}\bigg)^r$

$=(-1)^r . ^{15}C_r x^{(60 -4r -3r)}$

$=(-1)^r . ^{15}C_r x^{(60 -7r)}$

$\textbf{Step 2: Find value of coefficient of term x}^{32}$

$\text{To obtain term of x}^{32}$

$\Rightarrow 60-7r=32$

$\Rightarrow$ $7r=28$

$\therefore$ $r=4$

$\text{For r=4,5th term is given by-}$

$T_5=T_{4+1} = (-1)^4.^{15}C_4 x^{(60-7 \times 4)} = ^{15}C_4 x^{32} .$

$\therefore$ $\text{Coefficient of } x^{32}$$=^{15}C_4$

$= \dfrac { 15 \times 14 \times 13 \times 12 }{ 4 \times 3 \times 2\times 1}$

$= 1365$

$\textbf{Therefore,value of coefficient of term x$^{32}$ is 1365}$