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.
 A: 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 $(a, b, c) = (-1, 0, 1)$.
A: Using Wolfram Alpha:
WA:
$\alpha=-22+15\sqrt[3]{3}+9\sqrt[3]{9}$ is a root of $f(x)=x^3 + 66 x^2 + 237 x - 32768$.
WA:
$f(x^3)=(x^3 - 6 x^2 + 21 x - 32) (x^6 + 6 x^5 + 15 x^4 + 62 x^3 + 249 x^2 + 672 x + 1024)$.
This suggests that the (real) cubic root of $\alpha$ is a root of $g(x)=x^3 - 6 x^2 + 21 x - 32$, because the other factor is irreducible and has degree 6 and we expect the answer to be in the cubic field $\mathbb Q(\sqrt[3]3)$.
WA:
the only real root of $g$ is $\beta= 2 - \sqrt[3]{3} + \sqrt[3]{9}$.
WA:
it easy to check that $\beta^3=\alpha$.
A: (Too long for a comment.)

this is the system of equations after raise power of 3
$$
\begin{align}
p(a,b,c) = a^3+3b^3+9c^3+18abc+22 &= 0
\\ q(a,b,c) = a^2b+3ac^2+3b^2c-5 &= 0
\\ r(a,b,c) = a^2c+ab^2+3bc^2-3 &= 0
\end{align}
$$

Not something you'd do by hand, but that's an algebraic system which can always be reduced to one single polynomial equation using resultants, and then checked for rational solutions using the rational root theorem.
In this case for example, with WA's assistance you get 1 & 2
$\to$ 3 with the only rational root $c=1\,$:
$$
\begin{align}
u(b,c) = \text{res}_a (p,q) &= -3 b^9 + 27 b^6 c^3 - 22 b^6 - 81 b^5 c + 648 b^3 c^6 - 66 b^3 c^3
\\ &\quad - 486 b^2 c^4 - 198 b^2 c - 243 b c^2 + 81 c^9 + 396 c^6 + 484 c^3 - 27
\\v(b,c) = \text{res}_a(q,r) &= 3 b^7 c - 5 b^5 - 18 b^4 c^4 + 9 b^3 c^2 + 15 b^2 c^3 + 9 b^2 + 27 b c^7
\\ &\quad - 30 b c - 27 c^5 + 25 c^2
\\ w(c) = \text{res}_b(u,v) &= -9 (c - 1) (c^2 + c + 1) (3 c^3 + 1) (9 c^3 - 8) (\small{\dots\text{<degree 63 factor>} \dots})
\end{align}
$$
