20. (2002 National High School Competition) Real numbers $a$, $b$, $c$ and a positive number $\lambda$ make $f(x)=x^{3}+a x^{2}+b \dot{x}+c$ have three real roots $x_{1}$, $x_{2}$, $x_{3}$, and satisfy
(1) $x_{2}-x_{1}=\lambda$;
(2) $x_{3}>\frac{1}{2}\left(x_{1}+x_{2}\right)$.
Find the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$.
20. Since $f(x)=f(x)-f\left(x_{3}\right)=\left(x-x_{3}\right)\left[x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b\right]$, therefore, $x_{1}, x_{2}$ are the roots of the equation
$x^{2}+\left(a+x_{3}\right) x+x_{3}^{2}+a x_{3}+b=0$. From (1) we get
$$
\left(a+x_{3}\right)^{2}-4\left(x_{3}^{2}+a x_{3}+b\right)=\lambda^{2},
$$

which is $3 x_{3}^{2}+2 a x_{3}+\lambda^{2}+4 b-a^{2}=0$.
From (2) we get
$$
x_{3}=\frac{1}{3}\left(-a+\sqrt{4 a^{2}-12 b-3 \lambda^{2}}\right) \text {, }
$$

and $4 a^{2}-12 b-3 \lambda^{2} \geqslant 0$.
It is easy to see that $f(x)=x^{3}+a x^{2}+b x+c=\left(x+\frac{a}{3}\right)^{3}-\left(\frac{a^{2}}{3}-b\right)\left(x+\frac{a}{3}\right)+\frac{2}{27} a^{3}+c-\frac{1}{3} a b$.
From $f\left(x_{3}\right)=0$ we get
$$
\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\left(x_{3}+\frac{a}{3}\right)^{3}-\left(\frac{a^{2}}{3}-b\right)\left(x_{3}+\frac{a}{3}\right) \text {. }
$$

From (1) we get
$$
x_{3}+\frac{a}{3}=\frac{1}{3} \sqrt{4 a^{2}-12 b-3 \lambda^{2}}=\frac{2 \sqrt{3}}{3} \sqrt{\frac{a^{2}}{3}-b-\frac{\lambda^{2}}{4}} \text {. }
$$

Let $p=\frac{a^{2}}{3}-b$. From (2) and (3) we know $p \geqslant \frac{\lambda^{2}}{4}$ and
$\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} \sqrt{p-\frac{\lambda^{2}}{4}}\left(p-\lambda^{2}\right)$.
Let $y=\sqrt{p-\frac{\lambda^{2}}{4}}$, then $y \geqslant 0$ and
$\frac{1}{3} a b-\frac{2}{27} a^{3}-c=\frac{2 \sqrt{3}}{9} y\left(y^{2}-\frac{3}{4} \lambda^{2}\right)$.
Since $y^{3}-\frac{3 \lambda^{2}}{4} y+\frac{\lambda^{3}}{4}=y^{3}-\frac{3 \lambda^{2}}{4} y-\left(\frac{\lambda}{2}\right)^{3}+\frac{3 \lambda^{2}}{4} \cdot \frac{\lambda}{2}=\left(y-\frac{\lambda}{2}\right)^{2}(y+\lambda) \geqslant 0$,
then $\frac{1}{3} a b-\frac{2}{27} a^{3}-c-\frac{\sqrt{3}}{18} \lambda^{3}$.
Thus, $2 a^{3}+27 c-9 a b \leqslant \frac{3 \sqrt{3}}{2} \lambda^{3}$.
Therefore, $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}} \leqslant \frac{3 \sqrt{3}}{2}$.
Taking $a=2 \sqrt{3}, b=2, c=0, \lambda=2$, then $f(x)=x^{3}+2 \sqrt{3} x^{2}+2 x$ has roots $-\sqrt{3}-1,-\sqrt{3}+1$, 0. Clearly, the assumption holds, and
$$
\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}=\frac{1}{8}(48 \sqrt{3}-36 \sqrt{3})=\frac{3 \sqrt{3}}{2} .
$$

In summary, the maximum value of $\frac{2 a^{3}+27 c-9 a b}{\lambda^{3}}$ is $\frac{3 \sqrt{3}}{2}$.