# Computing the Minimal Polynomial of $\sqrt{3} + \sqrt[3]{5}$ over $\mathbb{Q}$

In trying to compute the minimal polynomial of $$\sqrt{3} + \sqrt[3]{5}$$ over $$\mathbb{Q}$$ I employed the usual approach of considering:

$$x - (\sqrt{3} + \sqrt[3]{5}$$) = 0

Then from there taking the constant term to the right-hand side, cubing both sides to then be left with:

$$x^3 - \sqrt{3}(3+5^{\frac{2}{3}}+ 2\times 5^{\frac{2}{3}}) + \sqrt[3]{5}(12 +5^3 +3)$$

The initial purpose of the problem was to show that the former number was algebraic over $$\mathbb{Q}$$, this is done in the usual fashion by first computing the minimal polynomial and then from there showing that indeed the coefficients are within that field, however, unless i've messed up somewhere in the computation, it doesn't seem to be that the constant term $$\sqrt{3}(3+5^{\frac{2}{3}}+ 2\times 5^{\frac{2}{3}}) + \sqrt[3]{5}(12 +5^3 +3) \in \mathbb{Q}$$

What this leads me to believe than, is that the former strategy perhaps, doesn't always work, therefore I would like to know if either I messed up on the computation or if there are any other methods for computing the minimal polynomial.

Any help would be appreciated.

Taking the entire constant term to the right side and cubing isn't the fastest way because the entire constant term is not a cube root. Instead, move only the cube root, and see what you get: $$x - \sqrt3 = \sqrt[3]{5}\\ (x - \sqrt3)^3 = \sqrt[3]{5}^3\\ x^3 - 3\sqrt3x^2 + 9x - 3\sqrt3=5$$ Now move all the square root terms to one side and square: $$x^3 +9x - 5 =\sqrt3(3x^2+3)\\ (x^3 +9x - 5)^2=(\sqrt3(3x^2+3))^2\\ x^6 +18x^4 -10x^3 + 81x^2 -90x + 25=27x^2+54x+27$$ Some tidying up should lead you to the polynomial you are after. Note that this approach is of limited use when you have more radicals, but in this case it barely got us to an answer.
Of course, while $$\sqrt3 + \sqrt[3]{5}$$ is a root of this polynomial, there is still the matter of proving that it's minimal. Eisenstein with $$p=2$$ does the trick, but you could also try to show that $$[\Bbb Q(\sqrt3 + \sqrt[3]{5}):\Bbb Q] = 6$$.
• I feel extremely unhappy because I know you are right but when computing (again and again) the resultant of $P(t)=t^3-5$ and $Q(t)=(x-t)^2-3$ (as a $5\times5$ determinant), I find $x^6-9x^4-10x^3+27x^2-90x-2$. Commented Oct 5, 2022 at 10:40