Find th eminimal polynomial of sqrt2+sqrt3 over QQ

To find the smallest degree polynomial f(x) with coefficients in Q, such that ${\displaystyle f(\sqrt{2}+\sqrt{3})=0.}$
The impo
rtant point is that the coefficients should all be in Q. For example, ${\displaystyle x-(\sqrt{2}+\sqrt{3})}$ is not allowed, as its coefficients are not rational numbers.
Now, ${\displaystyle \sqrt{3}\notin \mathbb{Q}\sqrt{2}.}$
So, we expect the minimal polynomial of ${\displaystyle \sqrt{2}+\sqrt{3}}$ to have degree ${\displaystyle 4({,}^{\prime }\sqrt{2}+\sqrt{3}\in \mathbb{Q}\left(\sqrt{2}\right)\cdot \left(\sqrt{3}\right)}$, a quadrantic extension of a quadrantic extension of ${\displaystyle \mathbb{Q}}$.
We proceed to find this minimal polynomial by repeated squaring (to clear radicals and obtain rational coefficients)
${\displaystyle x=\sqrt{2}+\sqrt{3}\Rightarrow x-\sqrt{2}=\sqrt{3},}$ squaring
${\displaystyle {(x-\sqrt{2})}^2=3\Rightarrow x^2-2\sqrt{2}x+2=3}$, rearrange
${\displaystyle x^2-1=2\sqrt{2}x}$, square again
${\displaystyle {(x^2-1)}^2=8x^2\Rightarrow x^4-2x^2+1=8x^2}$, rearrange
Minimal polynomial is ${\displaystyle x^4-10x^2+1}$