<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] 
But what if I didn't know about the cubic equation?

As long as you have a reasonable suspicion that the expression might be rational, the polynomial it satisfies can be derived rather easily. Let:
$$
x = \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}
$$
Then using $(a+b)^3=a^3+b^3+3ab(a+b)\,$ and $(56+19\sqrt{5})(56-19\sqrt{5})=1331=11^3\,$:
$$
\require{cancel}
\begin{align}
x^3 & = 19\sqrt{5} + 56 + \frac{11^3}{19\sqrt{5} + 56} + 3 \cdot \cancel{\sqrt[3]{19\sqrt{5} + 56}} \cdot \frac{11}{\cancel{\sqrt[3]{19\sqrt{5} + 56}}} \cdot x \\
 & = 56+ \cancel{19\sqrt{5}} + 56 - \cancel{19\sqrt{5}} + 33x \\
 & = 112 + 33x 
\end{align}
$$
By the rational root theorem, the equation $x^3-33x-112=0$ can only have rational roots that are integer divisors of $112=2^4\cdot 7\,$. Obviously $\pm 1$ do not satisfy the equation, and even roots can be eliminated by comparing the powers of $2$ between the terms, which leaves $\pm 7\,$ to try. 
</s>