<s>[INST] How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calculating the single real solution for the cubic equation $$ z^3 - 33 z - 112 = 0 $$
Using that cubic equation, I can show that $ z = 7 $ satisfies it and that a standard way of solving cubic equations shows that there is only a single real root and that its value is equal to the complicated expression, which is a very roundabout way of proving that the complicated expression is equal to 7.  But what if I didn't know about the cubic equation?  Is there a more direct way of reducing the complicated expression to a simpler one?
 [/INST] Using $a=\sqrt[3]{19\sqrt5+56}$ and $b=\frac{11}{\sqrt[3]{19\sqrt5+56}}$ and $u=a+b$, we have
$$
\begin{align}
a^3+b^3
&=19\sqrt5+56+\frac{1331}{19\sqrt5+56}\\
&=\frac{6272+2128\sqrt5}{56+19\sqrt5}\\[6pt]
&=112
\end{align}
$$
Thus,
$$
\begin{align}
&112\\
&=a^3+b^3\\
&=(a+b)\left(a^2-ab+b^2\right)\\
&=(a+b)\left((a+b)^2-3ab\right)\\
&=u(u^2-33)
\end{align}
$$
Noting that for $x\lt\sqrt{11}$,
$$
\begin{align}
x^3-33x-112
&\le22\sqrt{11}-112\\
&\lt0
\end{align}
$$
and for $x\ge\sqrt{11}$, $x^3-33x-112$ is increasing, we see that
$x^3-33x-112=0$ has only one real solution, and that is $x=7$.
Therefore, $\sqrt[3]{19\sqrt5+56}+\frac{11}{\sqrt[3]{19\sqrt5+56}}=a+b=u=7$.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                           
</s>