Answer
$495$
Work Step by Step
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 $x^0$:
$\binom{12}{8}=\dfrac{12!}{8!4!}=495$
