Uniform Continuity of $f(x) = x^3$ on $(0,1)$

Question

Trying to show that of $f(x) = x^3$ is uniformly continuous on (0,1).

Let $\epsilon > 0$. Now since $|x^3-y^3| = |x-y||x^2+xy+y^2| \leq |x-y||x^2+2|x||y|+y^2|$
$= |x-y|$ $||x|^2+2|x||y| + |y|^2|$ = $|x-y|$ $||x| + |y||^2$.

Then $|y-x| < 1$ then $|y| < |x| + 1$ and $|x-y|$ $||x| + |y||^2 < |x-y|$ $|2|x|+1|^2$. Here is where I am stuck. Any hints appreciated.

Answer 1

We have: for $0 < x,y < 1 \implies x^2+xy+y^2 < 3 \implies |x^3-y^3| < 3|x-y|$, and this is enough to show $f$ uniform continuous on $(0,1)$.