Question

The third term in the expansion of $\left(\sqrt[5]{x}-\frac{4}{\sqrt[3]{y^{2}}}\right)^{7}$ is equal to

• A. $-\frac{2240 x^{4 / 5}}{y^{2}}$
• B. $\frac{336 x}{y^{4 / 3}}$
• C. $\frac{2240 x^{4 / 5}}{y^{2}}$
• D. $-\frac{336 x^{3 / 5}}{y^{5 / 3}}$

Answer 1

Answer: (B)

$\frac{336 x}{y^{4 / 3}}$

Answer 2

The general term in the binomial expansion of $(a+b)^n$ is $T_{k+1} = \binom{n}{k} a^{n-k} b^k$. For the given expression, $a=x^{1/5}$, $b=-4y^{-2/3}$, and $n=7$. For the third term, $k=2$. Thus, $T_3 = \binom{7}{2}(x^{1/5})^{7-2}(-4y^{-2/3})^2 = 21 \cdot x \cdot 16y^{-4/3} = 336xy^{-4/3}$.