The Cube Is Good, But The Square Is Not

$$ Let $x$ be the number, $x^3=x$ and $x^2\neq x$ . $$ $\begin{aligned} x^3&=x\\ x^3-x&=0\\ x(x^2-1)&=0\\ x(x-1)(x+1)&=0\\ x_1=0,\; x_2=1,&\text{and}\;x_3=-1. \end{aligned}$ $$ Therefore, only $x_3=-1$ that meets the condition because $(-1)^3=-1$ and $(-1)^2\neq -1$ .

It has been proven without any reasonable doubt. $$

-i^3 is also one number. cube of -i^3 is also -i^3 but it's square is -1

(-1)^3=-1 Equal (-1)^2=1 Not Equal

(-1)^3 = -1 but (-1)^2 = 1

i is also a number, and it would appear to be a solution.

x3 = x

x2 x

x3 - x = 0

x(x2 - 1) = 0

x(x + 1)(x - 1) = 0

x = 0 or 1 or -1

02 = 0

12 = 1

Thus, answer is -1