Help with epsilon delta proof $x^3$ is near $27$ when $x$ is near $3$ but $x$ is not equal to $3$.
So I have
$$0<|x-3|<\delta \implies 3-\delta<x<3+\delta$$
$$|x^3-27|<\epsilon \implies|(x-3)(x^2+3x+3^2)|<\epsilon$$
$$=(x^2+3x+9)|(x-3)|<\epsilon \implies28|(x-3)|<\epsilon$$
$$=|(x-3)|<\epsilon/28$$
How do I prove $x^2+3x+9$ is less than $28$?
My attempt
$|x-3|<\delta$
so $$|x^2+3x+9|=|x^2+3x-18+27|\leq|x^2+3x-18|+27 \\ =|(x-3)(x+6)|+27
=|(x-3)(x-3+9)|+27$$
I'm not really sure where to go from here.

You can continue by saying that $|(x-3)(x-3+9)|+27\le|x-3|(|x-3|+9)+27<\delta(\delta+9)+27$, 
so now you just have to make sure that $\delta(\delta+9)\le1$ by, say, choosing $\delta$ so that it satisfies $\delta\le\frac{1}{18}$ 
since then $\delta^2\le\frac{1}{2}$ and $9\delta\le\frac{1}{2}$
(along with any other conditions required to ensure that $|x^3-27|<\epsilon$).
