Is this trick valid in proving that $f:(0,\infty)\to\mathbb{R}$; $f(x)=x^3$ is continuous? I wish to show that for $f:(0,\infty)$, with $f(x) = x^3$, is continuous for all $a\in(0,\infty)$.
I wondered whether the following trick is valid?
Notice that $|x^3-a^3|=|x-a|\cdot|x^2+ax+a^2|$. Now since $a,x$ are strictly positive, surely $x^2+ax+a^2<x^2+2ax+a^2=(x+a)^2$. Then $|x^3-a^3|<|x-a|(x+a)^2<\delta(x+a)^2$. Taking $\delta<1$, we see that $x+a<2a+1$. 
Hence given $\epsilon>0$, take $\delta:=\min(1,\frac{\epsilon}{(2a+1)^2})$. Then for all $x\in(0,\infty)$ with $0<|x-a|<\delta$ we have $$|x^3-a^3|<\delta(x+a)^2<\delta(2a+1)^2\leq\epsilon.$$
Have I made a mistake or was this a valid 'trick' in this particular context?

*

*Since you are proving continuity, instead of $0<|x-a|<\delta$, you need $|x-a|<\delta$ in your proof. 

*all you need is a bound on the quantity $(x+a)^2$ for $x$ near $a$. 

*What you did is OK but I would rewrite it in a slightly different way as follows.

Let $a$ be a fixed real number. Let $\epsilon>0$. We want to find a $\delta>0$ so that for all $x>0$ with $|x-a|<\delta$,
$$
|x^3-a^3|<\epsilon.\tag{1}
$$
Observe that since $x,a>0$,
$$
|x^3-a^3|=|x-a|\cdot|x^2+ax+a^2|=|x-a|(x^2+ax+a^2)\le |x-a|(x+a)^2\tag{2}
$$
On the other hand, for $x>0$ with $|x-a|<1$,
$$
(x+a)^2=(x-a+a)^2\leq 2(x-a)^2 +2a^2\le 2+2a^2\tag{3}
$$ 
Taking $\delta=\min(1,\frac{\epsilon}{2(1+a^2)})$, we conclude from (2) and (3) that 
$$
|x^3-a^3|\leq\delta\cdot 2(1+a^2)\leq \epsilon.
$$

[Added:]
You do not need to write $|x^2+ax+a^2|=x^2+ax+a^2$. Simply use 
$$
|x^2+ax+a^2|\leq |x|^2+|ax|+a^2
$$
and you can easily bound each of the quantities on the right for $x$ with $|x-a|<1$.
Such approach would give you a proof of continuity of $x^3$ on $\mathbb{R}$.
