The coefficient of x¹⁸ in the expansion of ${\left(x^4-\frac{1}{x^3}\right)}^{15}$is

${\left({\mathrm{x}}^4-\frac{1}{{\mathrm{x}}^3}\right)}^{15}\Rightarrow \frac{15!}{\mathrm{p}!\mathrm{q}!}{\left({\mathrm{x}}^4\right)}^{\mathrm{p}}{\left(-\frac{1}{{\mathrm{x}}^3}\right)}^{\mathrm{q}}$

$\Rightarrow \frac{15!}{\mathrm{p}!\mathrm{q}!}\left({\mathrm{x}}^{4\mathrm{p}}\right)(-1{)}^{\mathrm{q}}{\mathrm{x}}^{-3\mathrm{q}}$
$4\mathrm{p}-3\mathrm{q}=18\Rightarrow \mathrm{p}+\mathrm{q}=15$

Solving these two equations

 p = 9, q = 6

Coefficient of ${\mathrm{x}}^{18}\text{ is }\frac{15!}{9!6!}(-1{)}^6$

$\Rightarrow \frac{15\times 14\times 13\times 12\times 11\times 10\times 9!}{9!\times 6\times 120}=5005$