In the expansion of ${\left(x^4-\frac{1}{x^3}\right)}^{15}$ , the coefficient of $x^{39}$  is

Given expansion  ${\left(x^4-\frac{1}{x^3}\right)}^{15}$ is,

 We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$be the expansion of ${(x+a)}^n)$

 $T_{r+1}={}^{15}{\text{C}}_r{\left(x^4\right)}^{15-r}{\left(-\frac{1}{x^3}\right)}^r$

$T_{r+1}={}^{15}{\text{C}}_r{\left(x^4\right)}^{15-r}{\left(-1\right)}^r{\left(x^{-3}\right)}^r$

  $T_{r+1}={}^{15}{\text{C}}_r{x}^{60-4r-3r}{(-1)}^r$

$T_{r+1}={}^{15}{\text{C}}_r{x}^{60-7r}{(-1)}^r\mathrm{.........................}\left(1\right)$

$x^{60-7r}$ compare with $x^{39}$  because of finding $r$

$\Rightarrow x^{60-7r}=x^{39}$

This will contain $x^{39}$  if $60-7r=39$

$\because r=3$ this value substitute in Eqn (1)

$T_{3+1}={}^{15}{\text{C}}_3{x}^{60-7\times 3}{(-1)}^3$

$T_4={-}^{15}{\text{C}}_3{x}^{39}$

$T_4=-\frac{15\times 14\times 13}{3\times 2\times 1}{x}^{39}$

$\therefore$   $T_{3+1}$  i.e. $T_4$  is the term containing $x^{39}$

$\therefore$  Required coefficient  $={(-1)}^3{}^{15}{\text{C}}_3$

      $=-\frac{15\times 14\times 13}{1\times 2\times 3}=-455$