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So (x^4-\frac{1}{x^3})^{15}

so through binomial expansion  for (r + 1) term  in (a+x)^n  we know  


similarly for  (x^4-\frac{1}{x^3})^{15}

so let us forget about the coefficient part (^nC_r)

so x^{32}=x^{60-4r}*\frac{1}{x^3r}

⇒ x^{32}=x^{60-7r}

so 32 = 60 - 7r

⇒ 7r = 28 ⇒ r = 4

so the coefficient is =  ^15C_4 = 1380 ANSWER