using the primitive element of biquadratic field

Let $m$ and $n$ be two distinct squarefree integers $\ne 1$.  We want to express their square roots as polynomials of

$\alpha :=\sqrt{m}+\sqrt{n}$(1)

with rational coefficients.

If $\alpha$ is cubed (http://planetmath.org/CubeOfANumber), the result no terms with $\sqrt{mn}$:

${\alpha }^3$$\mathrm{\hspace{0.33em}}={(\sqrt{m})}^3+3{(\sqrt{m})}^2\sqrt{n}+3\sqrt{m}{(\sqrt{n})}^2+{(\sqrt{n})}^3$

$\mathrm{\hspace{0.33em}}=m\sqrt{m}+3m\sqrt{n}+3n\sqrt{m}+n\sqrt{n}$

$\mathrm{\hspace{0.33em}}=(m+3n)\sqrt{m}+(3m+n)\sqrt{n}$

Thus, if we subtract from this the product $(3m+n)\alpha$, the $\sqrt{n}$ term vanishes:

$${\alpha }^3-(3m+n)\alpha =(-2m+2n)\sqrt{m}$$

Dividing this equation by $-2m+2n$ ($\ne 0$) yields

$\sqrt{m}={\displaystyle \frac{{\alpha }^3-(3m+n)\alpha }{2(-m+n)}}.$(2)

Similarly, we have

$\sqrt{n}={\displaystyle \frac{{\alpha }^3-(m+3n)\alpha }{2(m-n)}}.$(3)

The (2) and (3) may be interpreted as such polynomials as intended.

Multiplying the equations (2) and (3) we obtain a corresponding for the square root of $mn$ which also lies in the quartic field  $\mathbb{Q}(\sqrt{m},\sqrt{n})=\mathbb{Q}(\sqrt{m}+\sqrt{n})$:

$\sqrt{mn}={\displaystyle \frac{{\alpha }^6-4(m+n){\alpha }^4+(3m^2+10mn+3n^2){\alpha }^2}{4(-m^2+2mn-n^2)}}$

For example, in the special case  $m:=2,n:=3$  we have

$$\sqrt{2}=\frac{{\alpha }^3-9\alpha }{2},\sqrt{3}=-\frac{{\alpha }^3-11\alpha }{2},\sqrt{6}=\frac{-{\alpha }^6+20{\alpha }^4-99{\alpha }^2}{4}.$$

Remark.  The sum (1) of two square roots of positive squarefree integers is always irrational, since in the contrary case, the equation (3) would say that $\sqrt{n}$ would be rational; this has been proven impossible here (http://planetmath.org/SquareRootOf2IsIrrationalProof).