2002
Problem - 3276
Write $\sqrt[3]{2+5\sqrt{3+2\sqrt{2}}}$ in the form of $a+b\sqrt{2}$ where $a$ and $b$ are integers.
Firstly, $$\sqrt[3]{2+5\sqrt{3+2\sqrt{2}}}=\sqrt[3]{2+5(1+\sqrt{2})}=\sqrt[3]{7+5\sqrt{2}}$$
Now, $$(a+b\sqrt{2})^3 = (a^3+6ab^2) + (3a^2b + 2b^3)\sqrt{2}$$
Hence
$$
\left\{
\begin{array}{}
a^3 + 6ab^2 &=7\\
3a^2 + 2b^3 &=5
\end{array}
\right.
\implies (a, b) = (1, 1)
$$
$$\therefore\quad \sqrt[3]{2+5\sqrt{3+2\sqrt{2}}} =\boxed{1+\sqrt{2}}$$