33 Find the largest real number $\lambda$, such that for a real-coefficient polynomial $f(x)=x^{3}+a x^{2}+c$ with all roots being non-negative real numbers, if $x \geqslant 0$, then $f(x) \geqslant \lambda(x-a)^{3}$, and find the condition for equality.
33. Let the three roots of $f(x)$ be $\alpha, \beta, \gamma$, and assume $0 \leqslant \alpha \leqslant \beta \leqslant \gamma$, then $x-a=x+\alpha+$ $\beta+\gamma, f(x)=(x-\alpha)(x-\beta)(x-\gamma)$. (1) When $0 \leqslant x \leqslant \alpha$, we have $-f(x)=$ $(\alpha-x)(\beta-x)(\gamma-x) \leqslant\left(\frac{\alpha+\beta+\gamma-3 x}{3}\right)^{3} \leqslant \frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=\frac{1}{27}(x-$ $a)^{3}$, so $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$. The equality holds when $x=0, \alpha=\beta=\gamma$; (2) When $\alpha \leqslant x \leqslant \beta$ or $x>\gamma$, $f(x)=(x-\alpha)(x-\beta)(x-\gamma)>0>-\frac{1}{27}(x-a)^{3}$. (3) When $\beta \leqslant x \leqslant \gamma$, $-f(x)=(x-\alpha)(x-\beta)(\gamma-x) \leqslant\left(\frac{x+\gamma-\alpha-\beta}{3}\right)^{3} \leqslant$ $\frac{1}{27}(x+\alpha+\beta+\gamma)^{3}=\frac{1}{27}(x-a)^{3}$. So $f(x) \geqslant-\frac{1}{27}(x-a)^{3}$. The equality holds when $\alpha=\beta=0$, $\gamma=2 x$. In summary, the maximum value of $\lambda$ is $-\frac{1}{27}$.