Question

Find the term indecent of x in the expansion of (x^2-1/x)^6

Answer 1

By the binomial theorem,

$\displaystyle \left(x^2-\frac1x\right)^6 = \sum_{k=0}^6 \binom 6k (x^2)^{6-k} \left(-\frac1x\right)^k = \sum_{k=0}^6 \binom 6k (-1)^k x^{12-3k}$

I assume you meant to say "independent", not "indecent", meaning we're looking for the constant term in the expansion. This happens for k such that

12 - 3k = 0   ===>   3k = 12   ===>   k = 4

which corresponds to the constant coefficient

$\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}$