3. Given that $a$ is a constant, the equation in terms of $x$
$$
x^{3}+\left(2 a-a^{2}\right) x-2 a^{2}=0
$$

has only one real solution in the real number range. Then the range of values for $a$ is $\qquad$ .
3. $0 \leqslant a<8$.

The original equation is equivalent to
$$
\begin{array}{l}
\left(x^{3}-a^{2} x\right)+\left(2 a x-2 a^{2}\right)=0 \\
\Leftrightarrow x(x+a)(x-a)+2 a(x-a)=0 \\
\Leftrightarrow(x-a)\left(x^{2}+a x+2 a\right)=0 \\
\Leftrightarrow x=a \text { or } x^{2}+a x+2 a=0 .
\end{array}
$$
(1) If the equation $x^{2}+a x+2 a=0$ has no solution, then
$$
a^{2}-4 \times 1 \times 2 a<0 \text {. }
$$

Solving this gives $0<a<8$.
(2) If $x=a$ is a solution to the equation $x^{2}+a x+2 a=0$, then
$$
a^{2}+a \times a+2 a=0 \text {. }
$$

Solving this gives $a=0$ or -1.
When $a=0$, the solutions to the original equation are $x_{1}=x_{2}=x_{3}=0$.
When $a=-1$, the solutions to the original equation are $x_{1}=x_{2}=-1$, $x_{3}=2$, which does not meet the requirements.
Therefore, the range of values for $a$ is $0 \leqslant a<8$.