Exercise 2.3.4

Proof. $\text{∙}$ Suppose that the polynomial $f=x^3+b{x}^2+\mathit{cx}+d=(x-x_1)(x-x_2)(x-x_3)$ has one root which is the average of the other two. We choose a numbering of the roots such that

Then

Let $s=x_1+x_2,p=x_1{x}_2$. The preceding equations give

We eliminate $s,p$ from these equations :

So the coefficients $b,c,d$ verify

$\text{∙}$ Conversely, suppose that $b,c,d$ verify

Let $s=-\frac{2}{3}b,p=c-\frac{2}{9}{b}^2$. Then $b=-\frac{3}{2}s,c=p+\frac{2}{9}{b}^2=p+\frac{1}{2}{(\frac{2}{3}b)}^2=p+\frac{1}{2}{s}^2$ : (6) and (7) are valid.

By the equation (4),

So $s,p$ verify the system (1),(2),(3) :

Let $x_1,x_2$ the complex roots of $x^2-\mathit{sx}+p$. Then $x_1+x_2=s,x_1{x}_2=p$. Let $x_3=\frac{x_1+x_2}{2}=\frac{1}{2}s$. Then

Thus $x_1,x_2,x_3$ are the roots of $(x-x_1)(x-x_2)(x-x_3)=x^3-{\sigma }_1{x}^2+{\sigma }_{2}x-{\sigma }_3=x^3+b{x}^2+\mathit{cx}+d$, and $x_3=\frac{x_1+x_2}{2}$.

Conclusion : one of the roots of $x^3+b{x}^2+\mathit{cx}+d$ is the average of the other two iff $2b^3-9\mathit{bc}+27d=0.$ □