Constructive proof that algebraic numbers form a field

Hint $ $ Let $\,a=\sqrt2,\,b=\sqrt[3]3.\,$ $\Bbb Q(a,b)\cong \Bbb Q\langle1,b,b^2, a,ab,ab^2\rangle$ as a vector space. $\,x\mapsto (a\!+\!b)\,x\,$ is a linear map on this vector space. Compute its matrix, then apply Cayley-Hamilton, to compute a (characteristic) polynomial having $\,a\!+\!b\,$ as root (e.g. by computing a determinant).

Remark $\ $ Such "determinant tricks" generalize to arbitrary rings. More efficiently (and precisely), one can use Grobner bases to do the elimination (e.g. via ideal-contraction), or resultants.

Mechanically, we can use resultants to eliminate $\,z,y\,$ from $\,x = y+z,\ x^2=2,\ y^3 = 3.$ First eliminating $\,z\,$ we have ${\rm res}(x\!-\!y\!-\!z,z^2\!-\!2,z) = (x\!-\!y)^2\!-\!2,\,$ then eliminating $\,y\,$ we have $\,{\rm res}((x\!-\!y)^2\!-\!2,y^2\!-\!3,y) = x^6\!-\!6x^4\!-\!6x^3\!+\!12x^2\!-\!36x\!+\!1\,$ (e.g. computed via Alpha)

Alternatively we can contract the ideal $\,(x\!-\!y\!-\!z,z^2\!-\!2,y^3\!-\!3)\,$ to $\,\Bbb Q[x],\,$ using Grobner bases.

The response of @BillDubuque is probably the best general method, but in this case, I’d start with the minimal $\Bbb Q$-polynomial for $\root3\of3$, namely $X^3-3$, and pass from this to $a$ polynomial for $\root3\of3+\sqrt2$, namely $f(X)=(X-\sqrt2)^3-3$, I’ll let you multiply it out; and then multiply $f$ by its $\Bbb Q(\sqrt2)$-conjugate, that is replace $\sqrt2$ by $-\sqrt2$ wherever it appears in $f$. You obviously get a $\Bbb Q$-polynomial, of the right degree, namely $6$, of which your number is a root.