<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] I'll outline the basic method to simplify$$\sqrt[3]{56+19\sqrt{5}}+\dfrac {11}{\sqrt[3]{56+19\sqrt{5}}}\tag{1}$$
To simplify the nested radical, note that we have this general outline:$$\sqrt[n]{A+B\sqrt[m]{C}}=a+b\sqrt[m]{C}\tag{2}$$
So therefore, we have$$\begin{align*}\sqrt[3]{56+19\sqrt{5}} & =a+b\sqrt{5}\\56+19\sqrt{5} & =\underbrace{(a^3+15ab^2)}_{56}+\underbrace{(3a^2b+5b^3)}_{19}\sqrt{5}\end{align*}$$
So we get this system of equations:$$\begin{align*} & a^3+15ab^2=56\\ & 3a^2b+5b^3=19\end{align*}\tag{3}$$
$(3)$ has real solutions as $(a,b)=\left(\dfrac 72,\dfrac 12\right)$ so$$\sqrt[3]{56+19\sqrt{5}}=\dfrac {7+\sqrt5}2\tag4$$
Using $(4)$, and through some algebraic manipulations, we have$$\begin{align*}\color{blue}{\sqrt[3]{56+19\sqrt5}}+\dfrac {11}{\color{red}{\sqrt[3]{56+19\sqrt{5}}}} & =\color{blue}{\dfrac {7+\sqrt{5}}2}+\dfrac {11}{\color{red}{\frac {7+\sqrt{5}}2}}\\ & =\dfrac {7+\sqrt{5}}{2}+\dfrac {22}{7+\sqrt{5}}\\ & =\dfrac {7+\sqrt{5}}2+\dfrac {22(7-\sqrt5)}{44}\\ & =\dfrac {7+\sqrt5}2+\dfrac {7-\sqrt5}2\\ & =\boxed 7\end{align*}$$
Just like what you got!
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