<s>[INST] Simple solving Skanavi book exercise: $\sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}$ Simple way to solve this exercise
$$
x = \sqrt[3]{9+\sqrt{80}}+\sqrt[3]{9-\sqrt{80}}
$$
 [/INST] For a hint:
Note that if $a+b+c=0$ then $a^3+b^3+c^3=3abc$ - can you see how to use that here?

To prove, let $a,b,c$ be the roots of $x^3-px^2+qx-r=0$ so that $p=a+b+c; q=ab+bc+ca; r=abc$ then$$a^3-pa^2+qa-r=0$$$$b^3-pb^2+qb-r=0$$$$c^3-pc^2+qc-r=0$$
Add these equations to obtain:$$(a^3+b^3+c^3)-p(a^2+b^2+c^2)+q(a+b+c)-3r$$And substituting and rearranging we get $$(a^3+b^3+c^3)-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$

To follow the comment about solutions to the cubic, examine the equation $$x^3+3px+q=0$$ and set $x=u+v$ then $$(u+v)^3=3uv(u+v)+u^3+v^3$$ //Note that this is $(u+v)^3-u^3-v^3=3(-u)(-v)(u+v)$ - which is another form of the earlier identity used by lab b in his solution//
and this is equivalent to $$x^3-3uvx-u^3-v^3$$
If we set $p=-uv$ and $q=-u^3-v^3$, we have $u^3, v^3$ as roots of $$y^2+qy-p^3=0$$Using the quadratic formula and $x=u+v$ we recover the form of the roots which is originally given.
Here $p=1, q=-18$ and the roots of the quadratic are $$\frac {18\pm\sqrt{324+-4}}{2}=9\pm \sqrt{80}$$
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