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?

Given $\epsilon>0$, we need $\delta>0$ such that if $0<|x-c|<\delta$, then $|x^3-c^3|<\epsilon$. Now,
$$
|x^3-c^3| = |x-c||x^2+xc+c^2|.
$$
If $|x-c|<1$, then we have that $-1<x-c<1$ or simply $c-1<x<c+1$ so that
$$
x^2+xc+c^2<(c+1)^2+(c+1)(c)+c^2=(c^2+2c+1)+(c^2+c)+c^2=3c^2+3c+1,
$$
and so
$$
|x^3-c^3| = |x-c||x^2+xc+c^2|<(3c^2+3c+1)|x-c|.
$$
So if we take $\delta=\min\left\{1,\frac{\epsilon}{3c^2+3c+1}\right\}$, then $0<|x-c|<\delta$ implies that
$$
|x^3-c^3|=|x-c||x^2+xc+c^2|<\frac{\epsilon}{3c^2+3c+1}\cdot(3c^2+3c+1)=\epsilon.
$$
Thus, by the definition of a limit, we have that
$$
\lim_{x\to c}x^3=c^3. \blacksquare
$$
