If $m = 0$, then $k = 0$ and $P(x) = x^3 + nx$. So $P$ has three real distinct roots: $0, \sqrt{-n}$ and $-\sqrt{-n}$.
Now suppose $m \neq 0$. If $m > 0$, then $P(x)$ has one sign change and $P(-x)$ has two sign changes. By Descartes' rule of signs, $P$ has one positive root and zero or two negative roots. Since $P(0) = \frac{mn}{9}$ and $P(-m) = -\frac{8mn}{9}$ have opposite signs, the intermediate value theorem ensures that $P$ has a negative root. Therefore $P$ has two negative roots. By a similar argument, if $m < 0$, then $P$ has one negative root and two positive roots. Hence in both cases, $P$ has three real roots, two of which have different signs. So it suffices to show that $P$ has no double root.
Since $P$ is cubic, if it has a double root, then that root is a critical point. The critical points of $P$ are $\frac{-m+\sqrt{m^2-3n}}{3}$ and $\frac{-m-\sqrt{m^2-3n}}{3}$. Neither of these satisfy $P$. Indeed, by the second derivative test, these points give a global min and max for $P$. But $P$ is nonconstant, so they cannot be roots of $P$. So, $P$ has no double root. Therefore, $P$ has three distinct real roots, with exactly two roots of the same sign.
