Find the term of the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ containing $x$ in the second power.
$\left(r+1\right)^{th}$ term in the expansion of $\left(a+b\right)^n$ is $\;\;$ $T_{r+1} = C^n_r \; a^{n-r} \; b^r$
Here, $\;$ $n = 7, \; a = \sqrt[3]{x^{-2}} = x^{\frac{-2}{3}}, \; b = x$
Let the $\left(r+1\right)^{th}$ term in the expansion of $\left(\sqrt[3]{x^{-2}} + x\right)^7$ contain $x$ in the second power.
Now, $\left(r+1\right)^{th}$ term is
$\begin{aligned}
T_{r+1} & = C^7_r \; \left(x^{\frac{-2}{3}}\right)^{7-r} \; x^r \\\\
& = C^7_r \; x^{\frac{-14}{3} + \frac{2r}{3} + r} \\\\
& = C^7_r \; x^{\frac{5r-14}{3}}
\end{aligned}$
$\because \;$ the $\left(r+1\right)^{th}$ term contains $x$ in the second power,
$\implies$ $\dfrac{5r - 14}{3} = 2$
i.e. $\;$ $5r - 14 = 6$
i.e. $\;$ $5r = 20$
$\implies$ $r = 4$
$\therefore \;$ The required term is
$\begin{aligned}
T_{4+1} = T_5 & = C^7_4 \; x^{\frac{20 - 14}{3}} \\\\
& = \dfrac{7!}{4! \times 3!} x^2 \\\\
& = \dfrac{7 \times 6 \times 5 \times 4!}{4! \times 6} x^2 \\\\
& = 35 x^2
\end{aligned}$