Powers of Real Number

$x^2 < x \Rightarrow x < 1$ , but this step can only be done for $x > 0$ (otherwise the inequality sign has to be switched), so the condition is equivalent to $0 < x < 1$ .

However, for these values, $x^3 = x \cdot x^2 < x^2$ , so there are no solutions .

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By the way, the possible orderings are

$\begin{cases} x^3 < x < x^2, \text{for } < -1 \\ x < x^3 < x^2, \text{for } -1 < x < 0 \\ x^3 < x^2 < x, \text{for } 0 < x < 1 \\ x < x^2 < x^3, \text{for } 1 < x \end{cases}$

So, $x^2 < x < x^3$ and $x^2 < x^3 < x$ are not possible.

LHS of this inequality: $x^2 - x < 0 \Rightarrow x(x-1) < 0 \Rightarrow x \in (0,1)$ (i).

RHS of this inequality: $x^3 - x > 0 \Rightarrow x(x+1)(x-1) > 0 \Rightarrow x \in (-1,0) \cup (1, \infty).$ (ii).

There are NO intervals of intersection between (i) and (ii), so no real solution $x$ exists.