I need to find the values of x that satisfy the given equation, and then find the distance between them. To do that, I can rewrite the equation as $x-\sqrt[5]{16}=\pm 3$ and then solve for x in each case. If $x-\sqrt[5]{16}=3$, then $x=3+\sqrt[5]{16}$. If $x-\sqrt[5]{16}=-3$, then $x=-3+\sqrt[5]{16}$. The distance between these two values of x is the same as the absolute value of their difference, which is $|3+\sqrt[5]{16}-(-3+\sqrt[5]{16})|$. I can simplify this expression by canceling out the $\sqrt[5]{16}$ terms and getting $|3-(-3)|=|6|=6$. 
What is the length of the segment of the number line whose endpoints satisfy $|x-\sqrt[5]{16}|=3$?