11. (15 points) Find the minimum value of $g(x)=\frac{x^{3}+1}{x^{2}-x}(x>1)$.


11. Solution: Let the required minimum value be $a$. Since $g(2)=\frac{9}{2}$ and $g(x)=3+\frac{(x-1)^{3}+2}{x^{2}-x}>3$, it follows that $31$ always holds and there exists $x>1$ such that $f(x)=0$. Therefore, the minimum value of $f(x)$ on $(0,+\infty)$ is 0.
$f^{\prime}(x)=3 x^{2}-2 a x+a$. Setting $f^{\prime}(x)=0$, we can solve for $x=\frac{a+\sqrt{a(a-3)}}{3}$, which is the unique critical point of $f(x)$ in $(1,+\infty)$. Moreover, $f(x)$ is decreasing on $\left(1, \frac{a+\sqrt{a(a-3)}}{3}\right]$ and increasing on $\left[\frac{a+\sqrt{a(a-3)}}{3},+\infty\right)$. Hence, the minimum value of $f(x)$ on $(1,+\infty)$ is $f\left(\frac{a+\sqrt{a(a-3)}}{3}\right)$, leading to $f\left(\frac{a+\sqrt{a(a-3)}}{3}\right)=0$,

which implies $(a+\sqrt{a(a-3)})^{3}-3 a(a+\sqrt{a(a-3)})^{3}+9 a(a+\sqrt{a(a-3)})+27=0$
$\Leftrightarrow-2 a^{3}+9 a^{2}+27=2 a(a-3) \sqrt{a(a-3)}$
$\Leftrightarrow 4 a^{6}-36 a^{5}+81 a^{4}-108 a^{3}+27 \cdot 18 a^{2}+27^{2}=$ $4 a^{3}\left(a^{3}-9 a^{2}+27 a-27\right)$
$\Leftrightarrow a^{4}-18 a^{2}-27=0$.
Given $3<a \leqslant \frac{9}{2}$, we can solve for the unique solution $a=\sqrt{9+6 \sqrt{3}}$. Therefore, the minimum value of $g(x)$ is $\sqrt{9+6 \sqrt{3}}$.