jmc

Note that if $r$ is a root, then so is $-r,$ so the roots are of the form $p,$ $-p,$ $q,$ $-q,$ for some complex numbers $p$ and $q.$ Since there are only two distinct roots, at least two of these values must be equal.

If $p=-p,$ then $p=0$ is a root. Hence, setting $x=0,$ we must get 0. In other words, $a^3=0,$ so $a=0.$ But then the polynomial is $$x^4-x^2=x^2(x-1)(x+1)=0,$$so there are three roots. Hence, there are no solutions in this case.

Otherwise, $p=\pm q,$ so the roots are of the form $p,$ $p,$ $-p,$ $-p,$ and the quartic is $$(x-p{)}^2(x+p{)}^2=x^4-2p^2{x}^2+p^4.$$Matching coefficients, we get $-2p^2=a^2-1$ and $p^4=a^3.$ Then $p^2=\frac{1-a^2}{2},$ so $${\left(\frac{1-a^2}{2}\right)}^2=a^3.$$This simplifies to $a^4-4a^3-2a^2+1=0.$

Let $f(x)=x^4-4x^3-2x^2+1.$ Since $f(0.51)>0$ and $f(0.52)<0,$ there is one root in the interval $(0.51,0.52).$ Since $f(4.43)<0$ and $f(4.44)>0,$ there is another root in the interval $(4.43,4.44).$ Factoring out these roots, we are left with a quadratic whose coefficients are approximately $$x^2+0.95x+0.44=0.$$The discriminant is negative, so this quadratic has two distinct, nonreal complex roots. Therefore, all the roots of $a^4-4a^3-2a^2+1=0$ are distinct, and by Vieta's formulas, their sum is $\boxed{4}.$