Evaluating constant term

Question

Consider you are given following

$$\biggr (x-\dfrac{2}{x^2}\biggr )^6$$

I'm trying to evaluate the constant term. What I've done so far is given below

$$\sum^{6}_{n = 0} \binom{6}{r}x^{6-r}\times (-2)^6 \times x^{-12}$$

$$\sum^{6}_{n = 0} \binom{6}{r}x^{-6-r}\times (-2)^6 $$

$$-6-r = 0 \implies r = -6$$

I got a negative number. Where did I go wrong?

Regards

Answer 1

It should be $$\binom6rx^{6-r}(-2x^{-2})^r=?$$

Answer 2

EDIT:

Here is a simple observation:

Constant term will have no $x$. So in order to omit $x$ we must have, $\displaystyle\frac{x^r}{x^{2(6-r)}}=1\implies x^r=x^{12-2r}\implies r=4$

Again since $^nC_r=^nC_{n-r}$ so $r=2$ will be another solution and in the expansion summing the Binomial quantities for $r=2$ and $r=4$ we get the constant term as $60$.

Here is a from Mathematica.

Hope it works.