nb
we have $\displaystyle \left( x^2 + x^{-1} \right)^{12}$
using $\displaystyle (a + b)^n = \sum_{k = 0}^{n}{n \choose k}a^kb^{n - k}$ we have:
$\displaystyle \left( x^2 + \frac 1x \right)^{12} = \left( x^2 + x^{-1} \right)^{12} = \sum_{k = 0}^{12}{12 \choose k}x^{2k}x^{k - 12} = \sum_{k = 0}^{12}{12 \choose k}x^{3k - 12}$
now continue