To show that \(\displaystyle f(x)= x^3\) is continuous at x= a, you must show that
"Given any \(\displaystyle \epsilon> 0\), there exist \(\displaystyle \delta> 0\) such that if \(\displaystyle |x- a|< \delta\) then \(\displaystyle |x^3- a^3|< \epsilon\)."

It will help to recognize that \(\displaystyle |x^3- a^3|= |x- a||x^2+ ax+ a^2|\) so you will need to find an upper bound on \(\displaystyle |x^2+ ax+ a^2|\).