By the binomial theorem,
[tex]\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}[/tex]
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
[tex]\dbinom 64 (-1)^4 = \dfrac{6!}{4!(6-4)!} = \boxed{15}[/tex]
