My proof ɛ-δ Definition of a limit only disproves the limit... What have I done wrong? Whilst attempting to prove that
$$\lim_{x\to8} \sqrt[3]{(3x+3)} = 3 $$ 
I came up with the following as my proof:
$$\lim_{x\to8} \sqrt[3]{(3x+3)} = 3 ⇔ (∀ε>0)(∃δ>0)(∀x)|x-8|<δ⇒|\sqrt[3]{(3x+3)}-\sqrt[3]{(27)}|≤ε⇒$$
$$|\sqrt[3]{(3x-24)}|≤ε⇒|\sqrt[3]{(3(x-8))}|≤ε∴|\sqrt[3]{(3δ)}|≤ε□$$
But that's when I realized that even though 
$$If δ=9, \;then \;ε≥3∵(f(x_{0}-δ)≤y_{0}-ε)$$
that
$$If δ=9, \;then \;ε≥3※∵¬(f(x_{0}+δ)≤y_{0}+ε)$$
and thus my proof falls apart. Where did I make a mistake? (or maybe I am misunderstanding, and my proof is correct?)
Thank you in advance!
Note: I am not 100% certain that I used all of these mathematical symbols correctly...
EDIT: Thanks to the comments that the wonderful people of math.stackexchange.com have provided below, I have concluded that
$$|\sqrt[3]{a}-\sqrt[3]{b}| ≠ |\sqrt[3]{(a-b)}|$$
Even so, the expression could be true, but they are not always equal. Therefore, I must re-evaluate this problem, and I will update this question if I find another (correct) solution. 
If anyone knows how to find the correct proof, please let me know what it is! 
Thank you!
SECOND EDIT: I have refined my proof to this:
$$\lim_{x\to8} \sqrt[3]{(3x+3)} = 3 ⇔ (∀ε>0)(∃δ>0)(∀x)|x-8|<δ⇒|\sqrt[3]{(3x+3)}-\sqrt[3]{(27)}|≤ε⇒$$
$$|\sqrt[3]{(3x-24)}-\sqrt[3]{27}+\sqrt[3]{27}|≤ε⇒|\sqrt[3]{(3x-24)}|≤ε⇒$$
$$|\sqrt[3]{(3(x-8))}|≤ε∴|\sqrt[3]{(3δ)}|≤ε□$$
Yet I come out to the same answer. Unless 
$$\sqrt[3]{(3x+3)}-\sqrt[3]{27} ≠ \sqrt[3]{3x+3}-\sqrt[3]{27}+\sqrt[3]{27}$$
I have no idea what's still wrong about it! 
Again, thank you for helping!
THIRD EDIT: I figured it out! The attempt above has the same problem as the first, I just made the mistake in a different way. Here's the real answer, which actually works.
$$\lim_{x\to8} \sqrt[3]{(3x+3)} = 3 ⇔ (∀ε>0)(∃δ>0)(∀x)|x-8|<δ⇒|\sqrt[3]{(3x+3)}-3|≤ε⇒$$
$$|\sqrt[3]{3x-24+27}-3|≤ε⇒|\sqrt[3]{3(x-8)+27}-3|≤ε∴\sqrt[3]{3δ+27}-3≤ε□$$
And that's it! It works! Thank you! 
 A: As pointed out in the comments, 
your mistake was in going from $|\sqrt[3]{3x+3}-\sqrt[3]{27}|\le\epsilon$ to $|\sqrt[3]{3x-24}|\le\epsilon$.

One way to find a $\delta$ that will work for a given $\epsilon>0$ is to use
$\displaystyle\sqrt[3]{A}-\sqrt[3]{B}=\frac{A-B}{A^{2/3}+A^{1/3}B^{1/3}+B^{2/3}}\;\;$ with $A=3x+3$ and $B=27$
This gives $\displaystyle\big|\sqrt[3]{3x+3}-3\big|=\left|\frac{3x-24}{(3x+3)^{2/3}+(3x+3)^{1/3}3^{1/3}+3^{2/3}}\right|=\frac{3|x-8|}{|(3x+3)^{2/3}+(3x+3)^{1/3}3^{1/3}+3^{2/3}|}$.
If we choose $\delta\le9$, then $|x-8|<\delta\implies x>-1\implies 3x+3>0$, 
so then $\displaystyle\big|\sqrt[3]{3x+3}-3\big|<\frac{3|x-8|}{3^{2/3}}<3|x-8|$.
 Therefore we also need to choose $\delta$ so that $3\delta<\epsilon$.

An alternate approach would be to use
$\big|\sqrt[3]{3x+3}-3\big|<\epsilon\iff3-\epsilon<\sqrt[3]{3x+3}<3+\epsilon\iff (3-\epsilon)^3<3x+3<(3+\epsilon)^3\iff$
$27-27\epsilon+9\epsilon^2-\epsilon^3<3x+3<27+27\epsilon+9\epsilon^2+\epsilon^3\iff$
$24-27\epsilon+9\epsilon^2-\epsilon^3<3x<24+27\epsilon+9\epsilon^2+\epsilon^3\iff$
$8-9\epsilon+3\epsilon^2-\frac{\epsilon^3}{3}<x<8+9\epsilon+3\epsilon^2+\frac{\epsilon^3}{3}$
Now if we let $d_1=8-(8-9\epsilon+3\epsilon^2-\frac{\epsilon^3}{3})=9\epsilon-3\epsilon^2+\frac{\epsilon^3}{3}$ 
$\hspace{.6 in}$and $d_2=(8+9\epsilon+3\epsilon^2+\frac{\epsilon^3}{3})-8=9\epsilon+3\epsilon^2+\frac{\epsilon^3}{3}$,
then we can check that $d_2>d_1>0$; so letting $\delta=d_1=9\epsilon-3\epsilon^2+\frac{\epsilon^3}{3}$
gives a $\delta>0$ such that $0<|x-8|<\delta\implies |\sqrt[3]{3x+3}-3|<\epsilon$
