True, or False?

If $x^3 = y^3$ ( $x,y \in \mathbb{R})$ , then we obtain:

$x^3 - y^3 = (x-y)(x^2 + xy + y^2) = 0$

Clearly, the left-hand factor requires $x = y$ , but the right-hand factor shows that:

$x^2 + xy + y^2 = 0 \Rightarrow x = \frac{-y \pm \sqrt{y^2 - 4(1)(y^2)}}{2} = \frac{-y \pm \sqrt{-3y^2}}{2} = (\frac{-1 \pm \sqrt{3}i}{2}) \cdot y$ .

which is a contradiction. Thus, the answer is true for $x,y \in \mathbb{R}.$

Since $f(a)=a^3$ is a strictly increasing function, so $f(x)=f(y)$ implies $x=y$