3、Let $a$ be a real number, and the equation
$$
x^{4}+3 a x^{3}+a\left(1-5 a^{2}\right) x-3 a^{4}+a^{2}+1=0
$$

has real roots, and there are at most 2 distinct real roots, then the value of $a$ is $\qquad$
Solution: $\pm \frac{2 \sqrt{26}}{13}$.
Let $x=y-a$, substituting into the original equation gives
$$
y^{4}-a y^{3}-3 a^{2} y^{2}+a y+1=0
$$

Since $y \neq 0$, we have
$$
\begin{array}{c}
y^{2}+\frac{1}{y^{2}}-a\left(y-\frac{1}{y}\right)-3 a^{2}=0 \\
\Rightarrow\left(y-\frac{1}{y}\right)^{2}-a\left(y-\frac{1}{y}\right)-3 a^{2}+2=0
\end{array}
$$

For any $t \in \mathbb{R}$
$$
y-\frac{1}{y}=t \Rightarrow y^{2}-t y-1=0 \Rightarrow \Delta>0
$$

If the equation $t^{2}-a t-3 a^{2}+2=0$ has no real roots, then the original equation has no real roots. Since one real root of this equation corresponds to two different real roots of the original equation, we have $a^{2}-4\left(-3 a^{2}+2\right)=0 \Rightarrow a= \pm \frac{2 \sqrt{26}}{13}$.