Mudulus compainsets the solutions

$f(x) = x^3-3|x|+2 = \begin{cases} f_+(x) = x^3-3x+2 & \text{for } x \ge 0 \\ f_- (x) = x^3+3x+2 & \text{for } x < 0 \end{cases}$

For $x \ge 0$ ,

$\begin{aligned} x^3-3x+2 & = 0 \\ (x-1)(x^2+x-2) & = 0 \\ (x-1)^2(x+2) & = 0 \\ \implies x & = 1 \end{aligned}$

Solution $x=-2 < 0$ is unacceptable.

For $x < 0$ , $\implies f_-(x) = x^3+3x+2$ $\implies f'_- (x) = 3x^2 + 3 > 0$ for all $x$ . Therefore, $f(x)_-$ is an always increasing function and it has only one solution. Let us check if the solution $x < 0$ . We note that $f_-(-1) = - 2$ and $f_-(0^-) = 2$ , therefore the solution is $-1 < x < 0$ .

Therefore, $f(x) = x^3-3|x|+2$ has $\boxed{2}$ solutions.

$x^3=3|x|-2 \Rightarrow f(x)=g(x)$ Where $f(x)=x^3$ and $g(x)=3|x|-2$ whose graphs are known, and we can be sure there is a negative solution at least. For positive $x$ we can compare the tangent line to $f$ with slope $3$ , which is $y=3x-2$ , which exactly the same positive branch of $g$ , so there is only one solution for positive $x$ . Therefore there are $2$ reals solutions.