Example 3. The cubic equation $x^{3}-a x^{2}-a^{2} x+2 a^{2}+a=0$ has three distinct real roots. Find the range of positive real values of $\mathrm{a}$.
Let $y=f(x)=x^{3}-a x^{2}-a^{2} x+2 a^{2}+a$,
$$
\begin{array}{l}
\left.f^{\prime} x\right)=3 x^{2}-2 a x-a^{2}=(3 x+a)(x-a). \\
\text { By } f^{\prime}(x)=0 \text { we get } x_{1}=-\frac{a}{3}, x_{2}=a, \\
\because a>0, \therefore x_{1}<0, x_{2}>0. \\
\text { When } x \in(-\infty, -\frac{a}{3}), f^{\prime}(x)>0, \text { function } y \text { is increasing; } \\
\text { When } x \in(-\frac{a}{3}, a), f^{\prime}(x)<0, \text { function } y \text { is decreasing; } \\
\text { When } x \in(a, +\infty), f^{\prime}(x)>0, \text { function } y \text { is increasing. } \\
\therefore \text { when } x=-\frac{a}{3}, y \text { reaches a local maximum; when } x=a, y \text { reaches a local minimum. } \\
\text { To make } y_{\text {min}}<0, \text { we need } f(a)<0, \text { i.e., } \\
a^{3}-a^{3}-a^{3}+2 a^{2}+a<0, \text { which simplifies to } -a^{3}+2 a^{2}+a<0; \text { by (2) we get } \\
a^{2}-2 a-1>0, \\
\therefore[a-(1+\sqrt{2})][a+(\sqrt{2}-1)]>0. \\
\because a>0, \therefore a+(\sqrt{2}-1)>0.
\end{array}
$$

Therefore, the range of $a$ is $a>1+\sqrt{2}$.