Q:

Finding a general way to construct least degree polynomial having rational coefficient having irrational roots

Let p(x) be the least degree polynomial equation having $\sqrt[3]{7}$ + $\sqrt[3]{49}$ as one of it's roots,  Then product of all roots of p(x) is ?

Following from the 'irrational roots occurring in pair for quadratics'
eg: $ (x^2 -2)  = (x + \sqrt{2} ) ( x-  \sqrt{2})$
I hypothesize that cubic roots would occur in triplets ( I look for proof of this and maybe generalization)
taking the generic cubic,
$$ p = ax^3 + bx^2 + cx + d  $$
Evaluating this cubic at the root given would be a pain, but I assume we could get a system of two equations from equating rational and irrational part separately to 0.
Find the polynomial equation of the lowest degree with rational coefficients whose one root is.......?
I saw this question, but one thing bothered me, if we plug one of the factors of coefficent as 'x' we get a completely rational polynomial but not otherwise. My basic question was this trick swapping coefficent with 'x' only possible because roots were equivalent? Or could we do this if roots were not equal?
The reason I am very suspicious with this method,  suppose
$$ p = (x-a)^2 =  x^2 + a^2 -2ax$$
and, since [x=a] (? this won't work at non root points)
$$ p=x^2 +a^2 -2a^2 = (x-a)(x+a)$$?

A:

Here is an explanation using basics of number fields.
The number $x = \sqrt[3]7 + \sqrt[3]{49}$ lives in a cubic number field, namely $\Bbb Q[\sqrt[3]{7}]$. This field has three embeddings into $\Bbb C$: you can send $\sqrt[3]7$ to $\sqrt[3]7$, $\sqrt[3]7 \omega$ or $\sqrt[3]7 \omega^2$, where $\omega = \frac{-1+\sqrt 3 i}2$ is a primitive third root of unity.
This means that the three conjugates of $x$ are:$$\sqrt[3]7 + \sqrt[3]{49}, \sqrt[3]7\omega + \sqrt[3]{49}\omega^2, \sqrt[3]7 \omega^2 + \sqrt[3]{49} \omega.$$
They are the roots of the minimal polynomial that you are looking for.
It is easy to see that their product equals $56$.

