a^3-a^2=(a^2-a)^2 Then a^3-a^2=a^4+a^2-2a^3 3a^3=a^4+2a^2 (1). Dividing by a^2 we get quadratic equation a^2-3a+2=0 which solutions are 2 and 1. The other case is when we can't divide (1) by a^2 then a=0

$a^3 - a^2 = (a^2-a)^2$, or $a^2(a^2-3a+2) = 0$. Factoring further yields $\boxed{0, 1, 2}$ as solutions.

Let the number be $x$. Then $x^3-x^2=(x^2-x)(x^2-x)$. Note that $x=0$ works. Now dividing by $x^2$ gives $x-1=(x-1)^2$. Note that $x=1$ works. Now dividing by $x-1$ gives $1=x-1$, so $x=2$ works. The answer is hence $\boxed{0,1,2}$.