\section*{Problem 6 - 041116}

Without using a table or a slide rule, it is to be decided whether the number

\[
z=\sqrt[3]{1620+12 \cdot \sqrt{17457}}+\sqrt[3]{1620-12 \cdot \sqrt{17457}}
\]

is greater, smaller, or equal to 18.
}

With the abbreviations \(a=1620, b=12 \sqrt{17457}\), we have \(z=\sqrt[3]{a+b}+\sqrt[3]{a-b}\), thus

\[
\begin{gathered}
z^{3}=a+b+3(a+b)^{\frac{2}{3}}(a-b)^{\frac{1}{3}}+3(a+b)^{\frac{1}{3}}(a-b)^{\frac{2}{3}}+a-b=2 a+3 \sqrt[3]{a^{2}-b^{2}} \cdot z= \\
=3240+3 \sqrt[3]{110592} \cdot z=3240+144 z
\end{gathered}
\]

(since \(110592=2^{12} \cdot 3^{3}=\left(2^{4} \cdot 3\right)^{3}\), no calculator is needed for 110592.

Therefore, \(z\) is a real root of the polynomial

\[
p:=x^{3}-144 x-3240=(x-18)\left(x^{2}+18 x+180\right)=(x-18)(x+9+\sqrt{-99})(x+9-\sqrt{-99})
\]

Since 18 is the only real root of \(p\), \(z\) must be equal to 18.