Question

Find the coefficient of the x^9 in the expansion of (3x^4 − 1/x)^6

Answer 1

By the binomial theorem:

$\left(3x^4-\dfrac1x\right)^6=\displaystyle\sum_{k=0}^6 \binom6k \left(3x^4\right)^{6-k} \left(-\dfrac1x\right)^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

is the binomial coefficient. Now,

$\dbinom6k \left(3x^4\right)^{6-k} \left(-\dfrac1x\right)^k=\dbinom6k 3^{6-k} (-1)^k x^{24-4k} x^{-k}=\dbinom 6k 3^6 \left(-\dfrac13\right)^k x^{24-5k}$

and the x⁹ term occurs when 24 - 5k = 9, or k = 3, so the coefficient is

$\dbinom63 3^{6-3} (-1)^3=-\dfrac{6!}{(3!)^2} 3^3=\boxed{-540}$