Calculus, how to find the value of delta in Epsilon-delta limit proof

Problem: Show that $\lim_{x\to 4}x^3=64$.

My steps:

$|f(x)-L|<\epsilon$ so $|x^3-64|<\epsilon$
$|x^3|<64+\epsilon \rightarrow 64-\epsilon < x^3 < 64+\epsilon$
$\sqrt[3]{64-\epsilon}$\sqrt[3]{64-\epsilon}-4So $\delta<\sqrt[3]{64+\epsilon}-4$
Let $\epsilon>0$. Pick $\delta$ so that $\delta<\sqrt[3]{64+\epsilon}-4$.
Suppose $0<|x-4|<\delta$. Then $|x-4|<\sqrt[3]{64+\epsilon}-4$.
Thus, $|x|<\sqrt[3]{64+\epsilon}$ and cube to get $|x^3|<64+\epsilon$.
Then $|x^3-64|<\epsilon$ which completes the proof.

But this is what the answer said:

Let $\epsilon>0$. Pick $\delta$ so that $\delta<1$ and $\delta<\frac{\epsilon}{61}$.
Suppose $0<|x−4|<\delta$.
Then $4−\deltaCube to get $(4−\delta)^3 Expanding the right-side inequality, we get
$x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64$
      $=64+\epsilon$.
The other inequality is similar.

So my questions are:
1. How can I get $\delta<\frac{\epsilon}{61}$?
2. Why $x^3<\delta^3+12*\delta^2+48*\delta+64<\delta+12\delta+48\delta+64=64+\epsilon$?
3. When we need to pick $\delta=min(1,\frac{\epsilon}{61}$) in this case?
4. Does my proof has problems ?

I am just a newbie to calculus... Thanks in advance.

Answer

Well, one is not supposed to use inverse functions like cube roots in expression for $\delta$ otherwise this leads to a circular argument.

The typical approach is to analyze your goal namely $$|x^3-64|<\epsilon$$ which has to be achieved using a specific kind of means namely $0<|x-4|<\delta$. So one first tries to simplify our goal by rewriting it as $$|x-4||x^2+4x+16|<\epsilon$$ The first factor in LHS can be controlled using our means but the second factor needs more work. One just needs to bound this factor suitably (again using the same means). We note that if $0<|x-4|<1$ then $3

Now we have the obvious inequality $$|x^3-64|=|x-4||x^2+4x+16|<61|x-4|$$ provided that $0<|x-4|<1$. If the RHS of the above inequality is less than $\epsilon$ then the LHS will also be less than $\epsilon$ and hence our goal is achieved if we have $61|x-4|<\epsilon$ or $|x-4|<\epsilon/61$. Combining the above we see that if $0<|x-4|<\min(1,\epsilon /61)$ then $|x^3-64|<\epsilon$. And hence $\delta$ can be taken as $\min(1,\epsilon /61)$. In case you have read the last paragraph carefully you can immediately see that the $\delta$ can also be taken as $\min(2,\epsilon /76)$.

Such problems do not have a unique answer and most importantly they are not supposed to be solved via algebraic manipulation of inequalities.