How do I find the constant term from a given binomial without expanding the whole expression?

${\left(x - \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{12}$

See below for an idea:

Explanation:

The constant term in a binomial expansion is the one, if it exists, that has no variable terms within it.

This prior answer on Socratic has a great explanation on how to find the constant term:

Following the instructions in that answer, we first set up the general term:

${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r} {\left(- \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{r}$

and now simplifying:

${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r} {\left(- 1\right)}^{r} \left({x}^{- \frac{r}{3}}\right)$

${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - r - \left(\frac{r}{3}\right)} {\left(- 1\right)}^{r}$

${T}_{r + 1} = \left(\begin{matrix}12 \\ r\end{matrix}\right) {x}^{12 - \frac{4 r}{3}} {\left(- 1\right)}^{r}$

The constant term is the one that will have $x = 0$, so we have:

${x}^{12 - \frac{4 r}{3}} = {x}^{0}$

giving

$12 - \frac{4 r}{3} = 0$

$12 = \frac{4 r}{3}$

$4 r = 36 \implies r = 9$

Therefore, the 9th term in the expansion is the one with no $x$ terms within it:

$\left(\begin{matrix}12 \\ 9\end{matrix}\right) {x}^{3} {\left(- \frac{1}{x} ^ \left(\frac{1}{3}\right)\right)}^{9}$

and to show that (i.e. checking our work):

$220 \times {x}^{3} \times \left(- \frac{1}{x} ^ 3\right) = - 220$