74. Consider positive real constants $m, n$, satisfying $3 n^{2}>m^{2}$. Real numbers $a, b, c$ satisfy the conditions: $a+b+c=m, a^{2}+b^{2}+c^{2}=n^{2}$. Find the maximum and minimum values of the expression $P$. $P=a^{2} b+b^{2} c+c^{2} a \quad$ (Le Trung Kien, Vo Quoc Ba Can)
Solution: Let $a=x+\frac{m}{3}, b=y+\frac{m}{3}, c=z+\frac{m}{3}$. From the given conditions, we get $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=\frac{3 n^{2}-m^{2}}{3}$, then the expression P becomes
$$P=x^{2} y+y^{2} z+z^{2} x+\frac{m^{3}}{9}$$

Notice that
$$\begin{array}{l}
\sum_{c y c}\left(3 x \sqrt{\frac{2}{3 n^{2}-m^{2}}}-\frac{18 x y}{3 n^{2}-m^{2}}-1\right)^{2}=3+\frac{18}{3 n^{2}-m^{2}}\left(\sum_{c y c} x\right)^{2}+\frac{324}{\left(3 n^{2}-m^{2}\right)^{2}} \sum_{c y c} x^{2} y^{2} \\
-6 \sqrt{\frac{2}{3 n^{2}-m^{2}} \sum_{c y c}} x-54\left(\frac{2}{3 n^{2}-m^{2}}\right)^{\frac{3}{2}} \sum_{c y c} x^{2} y \\
=3+\frac{324}{\left(3 n^{2}-m^{2}\right)^{2}} \sum_{c y c} x^{2} y^{2}-54\left(\frac{2}{3 n^{2}-m^{2}}\right)^{\frac{3}{2}} \sum_{c y c} x^{2} y
\end{array}$$

Since $x+y+z=0$, we have $x y+y z+z x=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)=-\frac{3 n^{2}-m^{2}}{6}$. Therefore,
$$\sum_{c y c} x^{2} y^{2}=\left(\sum_{c y c} x y\right)^{2}-2 x y z \sum_{c y c} x=\left(\sum_{c y c} x y\right)^{2}=\frac{\left(3 n^{2}-m^{2}\right)^{2}}{36}$$

We have
$12-54\left(\frac{2}{3 n^{2}-m^{2}}\right)^{\frac{3}{2}} \sum_{c y c} x^{2} y \geq 0$ or in other words
$$\sum_{c y c} x^{2} y \leq \frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}$$

If we choose
$$x=\frac{\sqrt{2\left(3 n^{2}-m^{2}\right)}}{3} \cos \frac{2 \pi}{9}, y=\frac{\sqrt{2\left(3 n^{2}-m^{2}\right)}}{3} \cos \frac{4 \pi}{9}, z=\frac{\sqrt{2\left(3 n^{2}-m^{2}\right)}}{3} \cos \frac{8 \pi}{9},$$

then $\square$
$$P=\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}+\frac{m^{3}}{9}$$

So $\max P=\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}+\frac{m^{3}}{9}$.
Similarly, by examining the expression $\sum_{c y c}\left(3 x \sqrt{\frac{2}{3 n^{2}-m^{2}}}+\frac{18 x y}{3 n^{2}-m^{2}}+1\right)^{2}$, we easily get
$$\min P=-\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{\frac{3}{2}}-\frac{m^{3}}{9}$$