How many of 'em fits in well?

$\begin{aligned} x^3 + x & = 0\\ \\ x(x^2 + 1) & = 0\\ \\ x = 0 & (\text{or}) x^2 + 1 = 0\\ \\ x = 0 & (\text{or}) x^2 = -1\\ \\ \text{But for real number x}, x^2 & \geq 0\\ \\ \implies x^2 & \neq -1\\ \\ \implies x & = 0 \end{aligned}$

Thus only $\color{#3D99F6}{\boxed{1}}$ value satisfies this equation.

$x^3+x=0\\ x(x^2+1)=0\\ \implies x=0, x^2=-1\\ \therefore x=0, x=i \text{ are solutions to the expression}$

Here $x=0$ is real whereas $x=i$ is imaginary so it have 1 real solution