If $\sqrt[3]{4x^2}=4$, find all possible values of $x$ and list them from least to greatest.
To solve for $x$, I need to get rid of the cube root on the left side. I can do this by raising both sides of the equation to the third power, since that will undo the cube root. So, I get $(\sqrt[3]{4x^2})^3=4^3$, which simplifies to $4x^2=64$. Now, I want to isolate $x$, so I divide both sides by $4$ and get $x^2=16$. To get $x$, I need to take the square root of both sides, but I have to remember that there are two possible roots: positive and negative. So, I get $x=\pm\sqrt{16}$, which means $x=\pm 4$. The question asks for the possible values of $x$ from least to greatest, so I write them as $-4, 4$. 