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.

Find range of values
Find the range of values of the constant a at which the equation ${x}^{3}-3{a}^{2}x+2=0$ has 3 different real number roots.
I took the derivative and found that $x=-a,a$. Then I solved for $f\left(a\right)=0$ and $f\left(-a\right)=0$ 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?
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Simon Zhang
Step 1
You're on the wrong path. The points where ${f}^{\prime }=0$ 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.
Step 2
On the other hand, you know that ${f}^{\prime }\left(a\right)=0$ and ${f}^{\prime }\left(-a\right)=0$, and the graph's a cubic, i.e., it goes from lower left to upper right. If it happens that $f\left(-a\right)>0$ and $f\left(a\right)<0$, then by the intermediate value theorem, f must have three zeroes.
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Elias Heath
Step 1
Since ${f}^{\prime }\left(x\right)=3{x}^{2}-3{a}^{2}=3\left(x-a\right)\left(x+a\right)$ you see that f is increasing on $\left(-\mathrm{\infty },-|a|\right]$ and on $\left[|a|,+\mathrm{\infty }\right)$ and decreasing in $\left[-|a|,|a|\right]$. On every such interval you can have at most one solution to the equation $f\left(x\right)=0$, because of monotonicity. To have a solution in each interval you need the function to change sign on the extreme points of the interval.
Step 2
So to have three solutions you need $f\left(-|a|\right)>0$ and $f\left(|a|\right)<0$ which gives:
$2|a{|}^{3}+2>0\phantom{\rule{0ex}{0ex}}-2|a{|}^{3}+2<0$
which is true for $|a|>1$.