How to find a minimal polynomial (field theory) I was asked to find a minimal polynomial of $$\alpha = \frac{3\sqrt{5} - 2\sqrt{7} + \sqrt{35}}{1 - \sqrt{5} + \sqrt{7}}$$ over Q.
I'm not able to find it without the help of WolframAlpha, which says that the minimal polynomial of $\alpha$ is $$19x^4 - 156x^3 - 280x^2 + 2312x + 3596.$$ (Truely it is - $\alpha$ is a root of the above polynomial and the above polynomial is also irreducible over Q.)
Can anyone help me with this?
Thank you!

To begin, clear denominators:
$$(1 - \sqrt{5} + \sqrt{7}) \alpha = 3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35}$$
We need to make the coefficient of $\alpha$ rational, so use a difference-of-squares trick to get rid of the $\sqrt{7}$ on the LHS (i.e. multiply both sides by $1 - \sqrt{5} - \sqrt{7}$),
$$((1 - \sqrt{5})^2 - 7) \alpha = (3 \sqrt{5} - 2 \sqrt{7} + \sqrt{35})(1 - \sqrt{5} - \sqrt{7})$$
and after expanding and collecting like terms:
$$(1 + 2 \sqrt{5}) \alpha = 1 + 4 \sqrt{5} + 7 \sqrt{7}$$
Now do the same again to deal with the $\sqrt{5}$ on the LHS:
$$19 \alpha = 39 - 2 \sqrt{5} - 7 \sqrt{7} + 14 \sqrt{35}$$
Next, we have to deal with the irrational numbers on the RHS. First, we deal with $\sqrt{5}$ (and $\sqrt{35}$): move all the other terms over to the LHS, and square the resulting equation,
$$(19 \alpha - 39 + 7 \sqrt{7})^2 = (-2 + 14 \sqrt{7})^2 \cdot 5$$
which expands to this
$$361 \alpha^2 - 1482 \alpha + 266 \sqrt{7} \alpha + 1864 -546 \sqrt{7} = 6880 - 280 \sqrt{7}$$
To finish off, we deal with $\sqrt{7}$: put all multiples of $\sqrt{7}$ on the RHS and all others on the LHS, and then square the resulting equation:
$$(361 \alpha^2 - 1482 \alpha - 5016)^2 = (- 266 \alpha + 266)^2 \cdot 7$$
Note that $19$ divides all the coefficients, so we can cancel that common factor:
$$(19 \alpha^2 - 78 \alpha - 264)^2 = (-14 \alpha + 14)^2 \cdot 7$$
Finally, we obtain,
$$361 \alpha^4 - 2964 \alpha^3 - 3984 \alpha^2 + 41184 \alpha + 69696 = 1372 \alpha^2 - 2744 \alpha + 1372$$
which simplifies to the desired equation:
$$19 \alpha^4 - 156 \alpha^3 - 280 \alpha^2 + 2312 \alpha + 3596 = 0$$
