We have $x_1,\:x_2,\:x_3\:\in \:\mathbb{C},\:\:f=x^3+x^2+mx+m,\:m\in \mathbb{R}$. We need to find $m\in\mathbb{R}$ such that $|x_1|=|x_2|=|x_3|$.
Here is what I tried: $f=x^3+x^2+mx+m=(x^2+m)(x+1)$, so one of root is:
$x=-1$, and if we put the condition from statement it means $|x_1|=|x_2|=|-1|=1$. From Vieta's formula, for cubic polynomial: $x_1x_2x_3=-m$. Therefore:
for $x_3=-1$ we'll obtain $m=-x_1x_2x_3=x_1x_2=1$
for $x_3=1$ we'll obtain $m=-x_1x_2x_3=-1$
I was wrong somewhere? Exist another method to solve the problem?
Your solution is the good one, nevertheless I do not quite understand how you go from Vieta's formula to the other statement.
As you have remarked $f(-1)=0$ whatever $m$ might be so you will always have $x_3=-1$ (up to renumerotation of the roots). Now this impose $|x_1|=|x_2|=|x_3|=1$ but because :
$$(x-x_1)(x-x_2)(x+1)=f(x)=(x^2+m)(x+1) $$
One must have :
$$x^2-(x_1+x_2)x+x_1x_2=(x-x_1)(x-x_2)=x^2+m $$
Hence one has :
$$x_1+x_2=0\text{ and } x_1x_2=m $$
That is :
$$x_1=-x_2\text{ and } -x_1^2=m $$
Now I claim that $|x_1|=1$ if and only if $|m|=1$. And any choice of $m$ with $|m|=1$ will give you solutions :
$$m=1\Leftrightarrow x_1^2=-1\Leftrightarrow (x_1,x_2)=(i,-i)\text{ or }(-i,i) $$
$$m=-1\Leftrightarrow x_1^2=1\Leftrightarrow (x_1,x_2)=(1,-1)\text{ or }(-1,1) $$
Hence to fullfill those conditions one must have $m=\pm 1$ and any choice of $m\in\{\pm1\}$ allows you to construct a triplet solution.