\section*{Problem 3 - 011043}

Let

\[
s=\sqrt[3]{20+14 \sqrt{2}}+\sqrt[3]{20-14 \sqrt{2}}
\]

Calculate \(s_{2}\) and \(s_{3}\) and try to find a rational value for \(s\)! (The root values must not be replaced by approximate values.)





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}

We set \(s=u+v\) with \(u=\sqrt[3]{20+14 \sqrt{2}}\) and \(v=\sqrt[3]{20-14 \sqrt{2}}\). Then we have

\[
u^{3}+v^{3}=(20+14 \sqrt{2})+(20-14 \sqrt{2})=40
\]
and further, by the third binomial formula,

\[
u^{3} v^{3}=(20+14 \sqrt{2}) \cdot (20-14 \sqrt{2})=8
\]

so \(u v=2\). Furthermore, it always holds that

\[
(u+v)^{3}=u^{3}+3 u^{2} v+3 u v^{2}+v^{3}=3 u v(u+v)+\left(u^{3}+v^{3}\right) \Rightarrow s^{3}-6 s-40=0
\]

Fortunately, a root of this cubic equation (which is already in reduced form) can be easily found by trial, namely \(s=4\).

A polynomial division by \((s-4)\) yields \(s^{3}-6 s-40=(s-4)\left(s^{2}+4 s+10\right)=0\), i.e., the other two roots are complex: \(s_{2,3}=-2+\sqrt{6} i\). The sought rational value for \(s\) is thus 4; the further powers are therefore \(s^{2}=16\) and \(s^{3}=64\).