<s>[INST] Denesting radicals $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$ and $\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$ I am trying to do denesting radicals:$$\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}$$ and $$\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}}$$
I tried to find Ramanujan polynomial like this link denesting radicals
But it doesn't work. I also tried to solve the system of equation $\sqrt[3]{-22+15\sqrt[3]{3}+9\sqrt[3]{9}}=a+b\sqrt[3]{3}+c\sqrt[3]{9}$ but it led to a scary-looking one.
 [/INST] Let $\mathbb{Q}(\sqrt[3]{3})$ be the set of all numbers of the form $a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for $a, b, c \in \mathbb{Q}$, and define the function $N(\cdot)$ on $\mathbb{Q}(\sqrt[3]{3})$ by
\begin{align*}
N(a + b\sqrt[3]{3} + c\sqrt[3]{9})
&= \prod_{k=0}^{2}(a + b \sqrt[3]{3}\,e^{2\pi i k/3} + c \sqrt[3]{9} \,e^{4\pi i k/3}) \\
&= a^3 + 3b^3 + 9c^3 - 9abc,
\end{align*}
Then it is not hard to check that, for $x, y \in \mathbb{Q}(\sqrt[3]{3})$,
$$ N(xy) = N(x)N(y). $$
Now, suppose that $x = \sqrt[3]{-22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}}$ is actually of the form $x = a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for some integers $a, b, c$. Then we have
\begin{align*}
N(x)^3 = N(x^3) &= N(-22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}) = 32768 = 2^{15}
\end{align*}
So, this and the equation $x^3 = -22 + 15\sqrt[3]{3} + 9\sqrt[3]{9}$ together yield
\begin{align*}
a^3 + 3b^3 + 9c^3 - 9abc &= N(x) = 32, \\
a^3 + 3b^3 + 9c^3 + 18abc &= -22, \\
3 a^2b + 9b^2 c + 9ac^2 &= -15, \\
3 ab^2 + 3a^2 c + 9bc^2 &= 9.
\end{align*}
From the first two equations, we get $27abc = -54$, or equivalently, $abc = -2$. This leaves only a handful of cases to be checked, and then it is not hard to find that $(a, b, c) = (2, -1, 1)$.
Generalizing this observation, we get:

Observation. Suppose that rational numbers $a, b, c, A, B, C$ satisfy the equation
$$ a + b\sqrt[3]{3} + c\sqrt[3]{9} = \sqrt[3]{A + B\sqrt[3]{3} + C\sqrt[3]{9}}. $$
Then we have
$$ 27abc = A - \sqrt[3]{A^3 + 3B^3 + 9C^3 - 9ABC}. $$


Likewise, assume that $\sqrt[3]{8-9\sqrt[3]{3}+3\sqrt[3]{9}} = a + b\sqrt[3]{3} + c\sqrt[3]{9}$ for some integers $a, b, c$. Then by the observation above, we get
$$ 27abc = 8 - \sqrt[3]{512} = 0 $$
and hence at least one of $a, b, c$ is zero. Indeed, when $b = 0$, we are led to solve the system of equations
\begin{align*}
a^3 + 9c^3 &= 8, &
9ac^2 &= -9, &
3a^2c &= 3.
\end{align*}
The last two equations shows that $c = -a$, and then the first equation gives $-8a^3 = 8$, hence $ a = -1$ and $c = 1$. Therefore $(a, b, c) = (-1, 0, 1)$.
</s>