Answer
$21ac^2\sqrt [5] {2a^2b^3}$
Work Step by Step
First, we have to make sure that the indices are the same, which they are, in this exercise.
Then, we rewrite each radicand as the product of perfect cubes and other factors:
$\sqrt [5] {-2^5 • 2a^5 • a^2b^3c^10} + 4ac\sqrt [5] {2ab^3c^5} + 19ac^2\sqrt [5] {2a^2b^3}$
Take the 5th root of terms:
$-2ac^2\sqrt [5] {2a^2b^3} + 4ac • c\sqrt [5] {2a^2b^3} + 19ac^2\sqrt [5] {2a^2b^3}$
Multiply coefficients:
$-2ac^2\sqrt [5] {2a^2b^3} + 4ac^2\sqrt [5] {2a^2b^3} + 19ac^2\sqrt [5] {2a^2b^3}$
Combine like terms:
$21ac^2\sqrt [5] {2a^2b^3}$