Proving $\lim_{x\to c}x^3=c^3$ for any $c\in\mathbb R$ using $\epsilon$-$\delta$ definition 
$\lim_{x\to 3}x^3=c^3$ for any $c\in\mathbb R$

Let $\epsilon>0$. Then
$$|x^3-c^3|=|x-c||x^2+xc+c^2|.$$
Let $\delta=\min\{1,\epsilon/x^2+xc+c^2\}$.
Then if $0<|x-c|<\delta$ and therefore since $|x-c|<\epsilon/(x^2+xc+c^2)$,
$$|x^3-c^3|=|x-c||x^2+xc+c^2|<ε.$$
Does this make sense or are the steps done in the right way?

Suppose $\varepsilon>0$. The inequality
$$
|x^3-c^3|<\varepsilon
$$
becomes
$$
c^3-\varepsilon<x^3<c^3+\varepsilon
$$
or
$$
\sqrt[3]{c^3-\varepsilon}<x<\sqrt[3]{c^3+\varepsilon}
$$
or also
$$
\sqrt[3]{c^3-\varepsilon}-c<x-c<\sqrt[3]{c^3+\varepsilon}-c
$$
Take
$$
\delta=\min\{c-\sqrt[3]{c^3-\varepsilon},\sqrt[3]{c^3+\varepsilon}-c\}
$$
after noting that
$$
c-\sqrt[3]{c^3-\varepsilon}>0
$$
and
$$
\sqrt[3]{c^3+\varepsilon}-c>0
$$
