11. Given the function $f(x)=\ln (a x+1)+x^{3}-x^{2}-a x$.
( I ) If $x=\frac{2}{3}$ is an extremum point of $y=f(x)$, find the value of the real number $a$;
(II) If $y=f(x)$ is an increasing function on $[1,+\infty)$, find the range of the real number $a$;
(III) If $a=-1$, the equation $f(1-x)-(1-x)^{3}=\frac{b}{x}$ has real roots, find the range of the real number $b$.
11. Analysis: (I) $f^{\prime}(x)=\frac{a}{a x+1}+3 x^{2}-2 x-a=\frac{x\left[3 a x^{2}+(3-2 a) x-\left(a^{2}+2\right)\right]}{a x+1}$

Since $x=\frac{2}{3}$ is an extremum point of $f(x)$, we have $f^{\prime}\left(\frac{2}{3}\right)=0$;

Thus, $3 a\left(\frac{2}{3}\right)^{2}+\frac{2}{3}(3-2 a)-\left(a^{2}+2\right)=0$, and $\frac{2}{3} a+1 \neq 0$, yielding $a=0$.

When $a=0$, $f^{\prime}(x)=x(3 x-2)$, so $x=\frac{2}{3}$ is indeed an extremum point of $f(x)$.
(II) Since $f(x)$ is an increasing function on $[1,+\infty)$,

we have $\frac{x\left[3 a x^{2}+(3-2 a) x-\left(a^{2}+2\right)\right]}{a x+1} \geq 0$ for all $x \in [1,+\infty)$.

If $a=0$, then $f^{\prime}(x)=x(3 x-2)$, and $f(x)$ is not an increasing function on $[1,+\infty)$;
If $a \neq 0$, since $a x+1>0$ for all $x>1$, we have $a>0$.

Thus, $3 a x^{2}+(3-2 a) x-\left(a^{2}+2\right) \geq 0$ for all $x \in [1,+\infty)$.
Let $g(x)=3 a x^{2}+(3-2 a) x-\left(a^{2}+2\right)$, with its axis of symmetry at $x=\frac{1}{3}-\frac{1}{2 a}$.
Since $a>0$, we have $\frac{1}{3}-\frac{1}{2 a}<0$, so $g(x)$ is increasing on $[1,+\infty)$.
Thus, $g(1) \geq 0$, which gives $3 a + (3 - 2a) - (a^2 + 2) \geq 0$, or $a^2 - a - 1 \leq 0$.
Solving this, we get $0 < a \leq \frac{1 + \sqrt{5}}{2}$.

Therefore, the range of $a$ is $\left(0, \frac{1 + \sqrt{5}}{2}\right]$.

(III) To find the range of $b$ such that the equation $x \ln x + x^2 - x^3 = b$ has a solution for $x > 0$,

we need to find the range of the function $g(x) = x \ln x + x^2 - x^3$.

Let $b = x (\ln x + x - x^2)$, and let $h(x) = \ln x + x - x^2$.
By $h^{\prime}(x) = \frac{1}{x} + 1 - 2x = \frac{(2x + 1)(1 - x)}{x}$, we have $h^{\prime}(x) > 0$ when $0 < x < 1$, so $h(x)$ is increasing on $(0,1)$; and $h^{\prime}(x) < 0$ when $x > 1$, so $h(x)$ is decreasing on $(1,+\infty)$.

Thus, $h(x) \leq h(1) = 0$, and $h(x)$ can be arbitrarily small. Therefore, the range of $b$ is $(-\infty, 0]$.