Find a polynomial in $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root

Question

What would be a way to find a polynomial in $\mathbb Q[x]$ such that $2^{1/2} + 3^{1/3}$ is a root.

Answer 1

We can say that $x=\sqrt[3]{2}+\sqrt[3]{3}$ and proceed from there.

$$x^3 = 5+ 3\sqrt[3]{12}+3\sqrt[3]{12}$$ $$x^3 -5 = 3(\sqrt[3]{12}+\sqrt[3]{18})$$ $$x^3 -5 = 3\sqrt[3]{6}(\sqrt[3]{2}+\sqrt[3]{3})$$ Now, since we already defined that $x=\sqrt[3]{2}+\sqrt[3]{3}$, we say that $$x^3 -5 = 3\sqrt[3]{6}(x)$$ $$x^3-3\sqrt[3]{6}x-5 = 0$$ And cubing, $$x^9-15x^3-87x^3-125$$

Answer 2

Think in terms of conjugates. $2^{1/2} + 3^{1/3}$ lives in the field $L := \Bbb{Q}(2^{1/2}, 3^{1/3}, \frac{-1+i\sqrt{3}}{2})$, so we could just take the product of all six linear factors of the form$$(x - a2^{1/2} - b3^{1/3}),$$ where $a = \pm 1$, $b = (\frac{-1 + i\sqrt{3}}{2})^k$ for $k = 0, 1, 2$. Since this polynomial's coefficients will be invariant under any automorphism $\sigma \in \operatorname{Gal}(L : \Bbb{Q})$, all the coefficients must actually live in $\Bbb{Q}$.

Answer 3

Usually what you want to do to find a polynomial such that a given $\alpha$ is a root, you take powers of $x-\alpha$ and play around with both sides of the equation. In this case:

$$ x-\alpha=0 \Longleftrightarrow x-\sqrt2=\sqrt[3]{3} $$ $$ (x-\sqrt2)^3 = 3 $$ You'll still have some radicals left here, but all you have to do is to move all of them on one side of the equation and take powers again of both sides, i.e. $$ x^3-3\sqrt2 x^2+6x-2\sqrt2 = 3 $$ $$ x^3+6x-3=\sqrt2 (3x^2+2) $$ $$ (x^3+6x-3)^2= 2(3x^2+2)^2, $$ which will finally give the degree-6 polynomial that you were looking for.