Three. (30 points) Given the equation
$$
\left(a x+a^{2}-1\right)^{2}+\frac{x^{2}}{(x+a)^{2}}+2 a^{2}-1=0
$$

has real solutions. Find the range of real values for $a$.
Three, obviously, $a \neq 0$. Given that the equation can be transformed into
$$
\left(a^{2} i x+a\right)^{2}-2 a x+\frac{x^{2}}{(x+a)^{2}}=0 \text {, }
$$

which is $\left[a(x+a)-\frac{x}{x+a}\right)^{2}=0$.
Thus, $a(x+a)-\frac{x}{x+a}=0$, then
$$
a \cdot x^{2}+\left(2 a^{2}-1\right) x+a^{3}=0 \text {. }
$$

By the discriminant, we get $-\frac{1}{2} \leqslant a \leqslant \frac{1}{2}$, but $a \neq 0$.