Question 77: Given non-negative real numbers $a$, $b$, $c$ satisfying $a+b+c=1$, try to find the maximum value of $a^{2}(b-c)+b^{2}(c-a)+$ $c^{2}(a-b)$.

保留源文本的换行和格式，翻译结果如下：

Question 77: Given non-negative real numbers $a$, $b$, $c$ satisfying $a+b+c=1$, try to find the maximum value of $a^{2}(b-c)+b^{2}(c-a)+$ $c^{2}(a-b)$.
Question 77, Solution: Notice that $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=(a-b)(a-c)(b-c)$. Without loss of generality, assume $a \geq b \geq c$, and let $a-b=x, b-c=y$, then $a+b+c=3 c+x+$ $2 y=1$. We need to find the maximum value of $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)=x y(x+y)$.

We only need to consider the case where $c=0$. In this case, $x=1-2 y$, so $x y(x+y)=(1-2 y)(1-y) y, y \in[0$, $\left.\frac{1}{2}\right]$.

Let $f(y)=(1-2 y)(1-y) y=2 y^{3}-3 y^{2}+y$, then $f(y)=6 y^{2}-6 y+1$, so $f(y)$ is monotonically increasing on $[0$, $\left.\frac{3-\sqrt{3}}{6}\right]$ and monotonically decreasing on $\left[\frac{3-\sqrt{3}}{6}, \frac{1}{2}\right]$. Therefore, we have:
$$
\max _{y \in\left[0, \frac{1}{2}\right]} f(y)=f\left(\frac{3-\sqrt{3}}{6}\right)=\frac{\sqrt{3}}{18}
$$

In conclusion, the maximum value of $a^{2}(b-c)+b^{2}(c-a)+c^{2}(a-b)$ is $\frac{\sqrt{3}}{18}$, which is achieved when $c=0, b=\frac{3-\sqrt{3}}{6}, a=$ $\frac{3+\sqrt{3}}{6}$.