## Precalculus (10th Edition)

$495$
We are given the expression: $\left(x^2+\dfrac{1}{x}\right)^{12}$ The term $T_{k+1}$ of the expansion of $(a+b)^n$ is: $T_{k+1}=\binom{n}{k}a^{n-k}b^r$ We have: $T_{k+1}=\binom{12}{k}(x^2)^{12-k}(x^{-1})^k=\binom{12}{k}x^{24-2k-k}=\binom{12}{k}x^{24-3k}$ Determine $k$ so that the term contains $x^0$: $24-3k=0$ $24=3k$ $k=8$ Determine the coefficient of $x^0$: $\binom{12}{8}=\dfrac{12!}{8!4!}=495$