SOLUTION. Let $a=x+\frac{m}{3}, b=y+\frac{m}{3}, c=z+\frac{m}{3}$. From the given conditions, we get that $x+y+z=0$ and $x^{2}+y^{2}+z^{2}=\frac{3 n^{2}-m^{2}}{3}$. The expression $P$ becomes
$$P=x^{2} y+y^{2} z+z^{2} x+\frac{m^{3}}{9}$$

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

Since $x+y+z=0$, we get $x y+y z+z x=-\frac{1}{2}\left(x^{2}+y^{2}+z^{2}\right)=-\frac{3 n^{2}-m^{2}}{6}$. Therefore
$$\sum_{c y c} x^{2} y^{2}=\left(\sum_{c y c} x y\right)^{2}-2 x y z \sum_{c y c} x=\left(\sum_{c y c} x y\right)^{2}=\frac{\left(3 n^{2}-m^{2}\right)^{2}}{36}$$
and we get
$$12-54\left(\frac{2}{3 n^{2}-m^{2}}\right)^{3 / 2} \sum_{c y c} x^{2} y \geq 0$$
or in other words,
$$\sum_{c y c} x^{2} y \leq \frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{3 / 2}$$

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

So
$$\max P=\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{3 / 2}+\frac{m^{3}}{9}$$

Similarly, by considering 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 conclude that
$$\min P=-\frac{2}{9}\left(\frac{3 n^{2}-m^{2}}{2}\right)^{3 / 2}-\frac{m^{3}}{9}$$

The problem is completely solved.