The value of $x=?$

Based on the given image, we have that $x^{3} - x = x^{2} - x^{3} \Longrightarrow x(x^{2} - 1) = x^{2}(1 - x) \Longrightarrow x(x - 1)(x + 1) = -x^{2}(x - 1)$ .

Now if $x$ were equal to either $0$ or $1$ then we would have $x = x^{2}$ , which is not the case here, so we can divide through by $x(x - 1)$ to find that

$x + 1 = -x \Longrightarrow 2x = -1 \Longrightarrow x = \boxed{-0.5}$ .

From the image, we deduce $\frac{x+x^2}{2}=x^3$

$\Rightarrow x+x^2=2x^3$

Since $x$ can't be zero:

$\Rightarrow 1+x=2x^2$

$\Rightarrow x=-\frac{1}{2} \; \vee \; x=1$

We also know that $x<x^3 <x^2$ , so must be $|x|<1$ and $x<0$ . Therefore $x=-\frac{1}{2}=\boxed{-0.5}$