Determine the minimal polynomial

Question

Determine the minimal polynomial of $v=\sqrt{3}+\sqrt[3]{2}$ over $\mathbb Q[x]$.

Can't find the right calculations. I am trying to find another way. I know the minimum polynomial of $w=\sqrt{3}+\sqrt{2}$ but still that does not help. I know it must be of degree 6 since the field extension of $\sqrt{3}$ + $\sqrt[3]{2}$ has degree 6.

Answer 1

Take the cube of both sides $v-\sqrt{3}=\sqrt[3]{2}$

$v^3-3v^2\sqrt{3}+9v-3\sqrt{3}=2$

We factorize: $3\sqrt{3}(v^2+1)=v^3+9v-2$.

A last squaring of both sides get rid of the last square root:

$27(v^4+2v^2+1)=v^6+81v^2+4+18v^4-36v-4v^3$

or:

$v^6 - 9\,v^4 - 4\,v^3 + 27\,v^2- 36\,v -23 = 0$

The left hand side is the minimal polynomial.