Conjugates of $12^{1/5}+54^{1/5}-144^{1/5}+648^{1/5}$ over $\mathbb{Q}$ After much manual computation I found the minimal polynomial to be $x^5+330x-4170$, although I would very much like to know if there's a clever way to see this. I suspect there is from seeing that the prime factorisations of the four integers are as various powers of $2$ and $3$ and because the number can be written as: $$12^{1/5}+54^{1/5}-(12^{1/5})^2+(12^{1/5}\cdot 54^{1/5})$$
But I haven't yet been able to find anything better than manual computation of the fifth power.
However, from here I am lost, I'm not sure how to solve this equation. A quick internet search returns much information on the general case in terms of the "Bring-Gerrard normal form" but the book from which this problem was taken hasn't gone into any detail on general methods for solving polynomials, so I am trying to find a solution that doesn't require any heavy machinery.
 A: Let $x$ be your number.
You can immediately see that $x \in \Bbb Q(12^{1/5}, 54^{1/5})$ (or even $x \in \Bbb Q(2^{1/5},3^{1/5})$, from your remark about the prime factors).
It is not too hard to show that its normal closure is $\Bbb Q(\zeta_5,12^{1/5},54^{1/5}) ( = \Bbb Q(\zeta_5,2^{1/5},3^{1/5}))$, and from there you can investigate the relevant Galois group and find the conjugates.
However, you have found that the number is actually of degree $5$, which should be a surprise at first : the situation is simpler than that.
After investigating a bit, we see that $54 = 12^3 / 2^5$, and so $54^{1/5}$ is already in $\Bbb Q(12^{1/5})$, and letting $y = 12^{1/5}$, we have $x = y + y^3/2 - y^2 + y^4/2$.
So now the problem reduces to finding the conjugates of $y = 12^{1/5}$. Its minimal polynomial is clearly $y^5-12=0$, and its $5$ roots are the the $\zeta_5^k 12^{1/5}$ where $\zeta_5$ is a primitive $5$th root of $1$. To get the conjugates of $x$ you simply replace $y$ with any of its conjugate in the above formula for $x$

In simpler terms : your calculations should actually show that "if $y^5 = 12$, then $(y+y^3/2-y^2+y^4/2)$ is a root of $x^5+330x-4170 = 0$". By using $5$ different $y$ such that $y^5 = 12$, you obtain $5$ roots of $x^5+330x-4170 = 0$. However, proving that they are distinct, and that the polynomial is irreducible, doesn't seem easy to do.
A: Noting that $12=\dfrac{18^5}{54^3}$ and $144 = \dfrac{6^5}{54}$, letting $\alpha = 54^{1/5}$, we see that the above is:
$$18\alpha^{-3} + \alpha - 6\alpha^{-1} + 18\alpha^{-2} = \frac{18\alpha^2 + 54\alpha - 6\alpha^4 + 18\alpha^3}{54}$$
Now replace $\alpha$ by conjugates of $\alpha$.
