$4 \cdot 37$ Try to point out, if the equation
$$x^{3}+a x^{2}+b x+c=0$$

has three real roots in arithmetic progression, what necessary and sufficient conditions should the real numbers $a, b, c$ satisfy?
［Solution］ Let the real-coefficient cubic equation be
$$x^{3}+a x^{2}+b x+c=0$$

with three real roots $x_{0}-y, x_{0}, x_{0}+y$ forming an arithmetic sequence. By the relationship between roots and coefficients, we have
$$\left(x_{0}-y\right)+x_{0}+\left(x_{0}+y\right)=-a,$$

so $x_{0}=-\frac{1}{3} a$.
Substituting $x = -\frac{1}{3} a$ into equation (1), we get
$$\left(-\frac{1}{3} a\right)^{3}+a\left(-\frac{1}{3} a\right)^{2}+b\left(-\frac{1}{3} a\right)+c=0$$

which simplifies to $2 a^{3}-9 a b+27 c=0$.
Conversely, if $a, b, c$ satisfy equation (2), then equation (1) has a root $x_{0}=-\frac{1}{3} a$. Let the other two roots of the equation be $x_{1}, x_{2}$. By the relationship between roots and coefficients, we have
$$x_{0}+x_{1}+x_{2}=-a, \quad x_{0} x_{1}+x_{1} x_{2}+x_{2} x_{0}=b,$$

and since $x_{0}=-\frac{1}{3} a$, we get
$$x_{1}+x_{2}=-\frac{2}{3} a, \quad x_{1} x_{2}=b-\frac{2}{9} a^{2}$$

Thus, $x_{1}, x_{0}, x_{2}$ form an arithmetic sequence, and the numbers $x_{1}$ and $x_{2}$ satisfy the quadratic equation
$$x^{2}+\frac{2}{3} a x+\left(b-\frac{2}{9} a^{2}\right)=0$$

Clearly, $x_{1}$ and $x_{2}$ are real numbers if and only if $\Delta=\left(\frac{2}{3} a\right)^{2}-4\left(b-\frac{2}{9} a^{2}\right) \geqslant 0$, which is equivalent to
$$a^{2}-3 b \geqslant 0$$

Therefore, the cubic equation (1) has three real roots forming an arithmetic sequence if and only if both conditions (2) and (3) are satisfied simultaneously.