Find the range of values of the constant $a$ at which the equation $x^3 - 3a^2x + 2 = 0$ has $3$ different real number roots.
I took the derivative and found that $x = -a, a$ Then I solved for $f(a) = 0$ and $f(-a) = 0$ to find that $a = -1, 1$
How do I use this information to find the range of values, or am I on the wrong path completely?
You're on the wrong path. The points where $f' = 0$ are "flat spots" in the graph. That doesn't tell you anything about where it's zero.
Actually, it tells you a little: between any two zeroes, there has to be a flat spot. So if $a = 0$, you can't possibly have three zeroes, because you've got only one flat spot, at 0.
On the other hand, you know that $f'(a) = 0$ and $f'(-a) = 0$, and the graph's a cubic, i.e., it goes from lower left to upper right. If it happens that $f(-a) > 0$ and $f(a) < 0$, then by the intermediate value theorem, $f$ must have three zeroes.
Since $$ f'(x) = 3x^2 -3a^2 = 3(x-a)(x+a) $$ you see that $f$ is increasing on $(-\infty,-|a|]$ and on $[|a|,+\infty)$ and decreasing in $[-|a|,|a|]$. On every such interval you can have at most one solution to the equation $f(x)=0$, because of monotonicity. To have a solution in each interval you need the function to change sign on the extreme points of the interval. So to have three solutions you need $f(-|a|)>0$ and $f(|a|)<0$ which gives: $$ 2|a|^3 + 2 >0\\ -2|a|^3 +2 <0 $$ which is true for $|a|>1$.
