What is the term independent of $x$ in the expansion of $(2x^{-1} + 3x^2)^{12}$? What is the term independent of $x$ in the expansion of $(2x^{-1} + 3x^2)^{12}$? 

I added the answer, is it asking what value is the expansion where x is not the coefficient? i.e the last answer in the picture?
 A: The $k$-th in the binomial expansion of this binomial is
$$\binom {12}k2^kx^{-k}3^{12-k}x^{2(12-k)}=\binom {12}k2^k3^{12-k}x^{24-3k},$$
and it is a constant if and only if $k=8$, in which case the coefficient is
$$\binom {12}82^8\,3^4=10\,264\,320.$$
A: Another approach getting Bernard's answer, using the coefficient-of operator:
$$
\begin{align}
\left[x^0\right]\left(2x^{-1}+3x^2\right)^{12}
&=\left[x^{12}\right]\left(2+3x^3\right)^{12}\\[6pt]
&=\left[x^4\right]\left(2+3x\right)^{12}\\
&=\binom{12}{4}\,2^8\,3^4\\
&=10264320
\end{align}
$$
A: The independent term is the term where the exponent of $x$ is zero. So yes, in this case, it would be the final number, 10 264 320.
A: The question is asking which term in that expansion is the coefficient of $x^0$, aka the constant coefficient. Which, in this case, it that last term.
However, I would guess that the question is not asking to test your Wolfram Alpha skills, so I would recommend understanding how you could find just that term without doing the full expansion (hint: put all the terms in the expansion in order of powers of $x$, and see how they jump; maybe you could factor something out of the binomial expression and then apply a familiar formula?)
