Trying to evaluate the constant term

Question

Suppose you are given that

$$\biggr (\sqrt[3]x - \dfrac{1}{\sqrt x} \biggr )^{15}$$

I'm trying to evaluate the constant term.

$$\sum^{15}_{n = 0} \binom{6}{r}x^{5-r}\cdot -\dfrac{\sqrt x}{x}$$

How can I take it from there?

Answer 1

$$\sum^{15}_{n = 0} \binom{6}{r}x^{5-r}\cdot -\dfrac{\sqrt x}{x}$$

Why $6$? This looks a bit strange; from the binomial theorem:

$$\biggr (\sqrt[3]x - \dfrac{1}{\sqrt x} \biggr )^{15} = \sum^{15}_{n = 0} \binom{15}{n}\left(\sqrt[3]x\right)^n\left(\dfrac{-1}{\sqrt x}\right)^{15-n}\tag{$*$}$$

Rewrite:

$$\begin{align} \left(\sqrt[3]x\right)^n\left(\dfrac{-1}{\sqrt x}\right)^{15-n}& =\left(-1\right)^{15-n}\left(x^\tfrac{1}{3}\right)^n\left(x^{-\tfrac{1}{2}}\right)^{15-n} \\[8pt] & =\left(-1\right)^{15-n}\;x^{\tfrac{n}{3}}x^{-\tfrac{1}{2}(15-n)} \\[8pt] & =\left(-1\right)^{15-n}\;x^{\color{blue}{\tfrac{n}{3}-\tfrac{1}{2}(15-n)}} \end{align}$$

You have the constant term if the exponent of $x$ is equal to $0$, so when: $${\color{blue}{\tfrac{n}{3}-\tfrac{1}{2}(15-n)}}=0 \iff n = \ldots$$ Plug this value for $n$ back in $(*)$ but it is clear the coefficient is simply $\left(-1\right)^{15-n}\binom{15}{n}$.