Find the term of the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ which does not contain $x$.
$\left(k + 1\right)^{th}$ term in the expansion of $\left(a + b\right)^n$ is \;\; $T_{k+1} = C^n_k \; a^{n-k} \; b^k$
Let the $\left(k + 1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15} = \left(x^{\frac{1}{3}} - x^{\frac{-1}{2}}\right)^{15}$ be independent of $x$.
Now, $\;$ $T_{k+1} = C^{15}_k \; \left(x^{\frac{1}{3}}\right)^{15 - k} \; \left(x^{\frac{-1}{2}}\right)^{k} = C^{15}_k \; x^{5- \frac{k}{3} - \frac{k}{2}}$
Since this term is independent of $x$ $\implies$ $5 - \dfrac{k}{3} - \dfrac{k}{2} = 0$
i.e. $\;$ $5 = \dfrac{5k}{6}$ $\implies$ $k = 6$
i.e. $7^{th}$ term is independent of $x$.
$\therefore \;$ The term independent of $x$ in the expansion of $\left(\sqrt[3]{x} - \dfrac{1}{\sqrt{x}}\right)^{15}$ is
$T_{7} = T_{6+1} = C^{15}_{6} = \dfrac{15!}{6! \left(15 - 6\right)!} = \dfrac{15!}{6! \times 9!} = 5005$