Answer
$\sqrt[5]{77x^3}$
Work Step by Step
Simplify. $\sqrt[5] {7x^2}\sqrt[5]{11x}$
Thus,
As per the product rule, we have $\sqrt[n] {pq}=\sqrt[n] {p}\sqrt[n] {q}$
$\sqrt[5] {7x^2}\sqrt[5]{11x}=\sqrt[5]{(7x^2)(11x)}=\sqrt[5]{(77)(x \cdot x^2)}$
Hence, the above exponent in radical form can be written as:
$\sqrt[5]{77x^3}$
