Find $m$ such that a given polynomial has all its roots real

Let $f = X^3 + mX^2 + mX + 1$ be a polynomial with real coefficients and $m \in \mathbb{R}$. Find $m$ such that all of $f$'s roots are real.

I could only think about having the following condition:
$$x_1^2 + x_2^2 + x_3^2 \geq 0$$

This way, I've got $m \in (-\infty, 0) \cup (2, \infty)$

Answer

Hint: Try $X=-1$, and thus factorize the cubic into quadratic polynomial.

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Answer

You'll get
$$X^3+mX^2+mX+1=(X+1)(X^2+(m-1)X+1)$$
using long division. Which means that
$$\Delta=(m-1)^2-4\ge 0$$
$$\implies m^2-2m-3\ge0$$
$$\implies m\ge 3 \;or\; m\le -1$$
And using the notation of set theory,
$$m\in (-\infty , -1]\cup [3,\infty )$$