<s>[INST] What is the value of $\sqrt[3]{9+2\sqrt5+3\sqrt{15}}$? 
$$\sqrt[3]{9+2\sqrt5+3\sqrt{15}}=?$$

Here is my work:
I considered the value is equal to $a+b$. Hence
$$9+2\sqrt5+3\sqrt{15}=(a+b)^3=a^3+b^3+3a^2b+3ab^2$$
$$2\sqrt5+3(3+\sqrt{15})=a^3+b^3+3(a^2b+ab^2)$$
We have
$$ \begin{cases} {a^2b+ab^2=3+\sqrt{15}} \\ {a^3+b^3=2\sqrt5} \end{cases}    $$
But I cannot find such $a$ and $b$.
 [/INST] Remark: Write $$9+2\sqrt{5}+3\sqrt{15}=(a+b\sqrt{3}+c\sqrt{5})^3.$$
Then the equations in rational $a,b,c$ yield immediately a contradiction.
Of course, there might be more complicated expressions. But for such questions, usually the answer is simple.
Edit: The equations are
\begin{align*}
0 & =6abc - 3\\
0 & = 3a^2c + 9b^2c + 5c^3 - 2\\
0 & =3a^2b + 3b^3 + 15bc^2\\
0 & =a^3 + 9ab^2 + 15ac^2 - 9.
\end{align*}
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