Simplification of expressions containing radicals As an example, consider the polynomial $f(x) = x^3 + x - 2 = (x - 1)(x^2 + x + 2)$ which clearly has a root $x = 1$.
But we can also find the roots using Cardano's method, which leads to
$$x = \sqrt[3]{\sqrt{28/27} + 1} - \sqrt[3]{\sqrt{28/27} - 1}$$
and two other roots.
It's easy to check numerically that this expression is really equal to $1$, but is there a way to derive it algebraically which isn't equivalent to showing that this expression satisfies $f(x) = 0$?

Pardon my skepticism, but has anyone so much as breadboarded Blömer '92 or Landau '93 in all these 18 years?  For lack of same, people still publish ugly surdballs, e.g.,
$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt[3]{-4+3 \sqrt{2}+3 \sqrt[4]{3}+2 \sqrt{3}-3^{3/4}+2 \sqrt{2}\, 3^{3/4}} \sqrt[4]{\pi }}{2\
   3^{3/8} \sqrt[6]{\left(\sqrt{2}-1\right) \left(\sqrt{3}-1\right)} \Gamma \left(\frac{3}{4}\right)}$$
(J. Yi / J. Math. Anal. Appl. 292 (2004) 381–400, Thm 5.5 vi) instead of
$$\vartheta _3\left(0,e^{-6 \pi }\right)=\frac{\sqrt{2+\sqrt{2}+\sqrt{2} \sqrt[4]{3}+\sqrt{6}} \,\sqrt[4]{\pi }}{2\ 3^{3/8} \Gamma
   \left(\frac{3}{4}\right)}\quad .$$
And why do both papers trot out the same old Ramanujan denestings instead of new and interesting ones?  E.g.,
$$\sqrt{2^{6/7}-1}=\frac{2^{8/7}-2^{6/7}+2^{5/7}+2^{3/7}-1}{\sqrt{7}}$$
or
$$\sqrt[3]{3^{3/5}-\sqrt[5]{2}}=\frac{2^{2/5}+\sqrt[5]{3}+2^{3/5} 3^{2/5}-\sqrt[5]{2}\, 3^{3/5}}{5^{2/3}}$$
or
$$\frac{\sqrt[3]{1+\sqrt{3}+\sqrt{2}\, 3^{3/4}}}{\sqrt[6]{\sqrt{3}-1}}=\frac{\sqrt{1+\sqrt{3}+\sqrt{2} \sqrt[4]{3}}}{\sqrt[6]{2}}\quad ?$$
These results were found by two young students of mine who would very much like to know values of q and b in Bill Dubuque's structure theorem which effect the denesting
$$\sqrt[3]{-\frac{106}{25}-\frac{369 \sqrt{3}}{125}+\frac{3 \sqrt{3} \left(388+268 \sqrt{3}\right)}{100 \sqrt[3]{2}\,
   5^{2/3}}}=\frac{3}{5^{2/3}}-\frac{1+\sqrt{3}}{\sqrt[3]{10}}+\frac{1}{5} \sqrt[3]{2} \left(3+2 \sqrt{3}\right)\quad.$$
Thanks in advance.
