Problem source

\begin{questionparts}
\item
Express $\left(3+2\sqrt{5} \, \right)^3$ 
in the form $a+b\sqrt{5}$ where $a$ and $b$ are integers.
\item Find the positive integers $c$ and $d$ such that 
$\sqrt[3]{99-70\sqrt{2}\;}$ =  $c - d\sqrt{2} \,$.
\item Find the two  real solutions of $x^6 - 198 x^3 + 1 = 0 \,$.
\end{questionparts}

Solution source

\begin{questionparts}

\item \begin{align*}
(3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\
&= 27 + 180 + (54+40)\sqrt{5} \\
&= 207 + 94\sqrt{5}
\end{align*}

\item \begin{align*}
&& (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\
\Rightarrow && 99 &= c(c^2+6d^2) \\
&& 70 &=  d(3c^2+2d^2) \\
\Rightarrow && c & \mid 99, d \mid 70 \\
&& c &= 3, d = 2
\end{align*}
\item \begin{align*}
&& 0 &= x^6 - 198x^3 + 1 \\
\Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\
\Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\
&&&= 99 \pm 10 \sqrt{98} \\
&&&= 99 \pm 70 \sqrt{2} \\
\Rightarrow && x &= 3 \pm 2 \sqrt{2}
\end{align*}
\end{questionparts}