Polynomial with root $α = \sqrt{2}+\sqrt{5}$ and using it to simplify $α^6$ Find a polynomial $\space P(X) \in \mathbb{Q}[X]\space$ of degree 4 such that
$$\alpha = \sqrt{2} + \sqrt{5}$$
Is a root of $P$. 
Using this polynomial, find numbers $\space a, b, c, d \space$ such that
$$\alpha^{6} = a + b\alpha + c\alpha^{2} + d\alpha^{3}$$
What have I tried so far? 
I know obviously that for $\alpha$ to be a root of $P$, then $(x-\alpha)$ must be part of the polynomial. Hence, $(x-\sqrt{2} - \sqrt{5})$ will be a factor of the polynomial. Where I’m getting stuck is what to do next in order to find the other factors of the polynomial such that I get values $a, b, c$ and $d$ that satisfy the equation with $\alpha^{6}$. 
Any help would be greatly appreciated!

We can use the following identity.
$$(a+b+c)\prod_{cyc}(a+b-c)=\sum_{cyc}(2a^2b^2-a^4).$$
We obtain:
$$2\alpha^2\cdot2+2\alpha^2\cdot5+2\cdot2\cdot5-\alpha^4-2^2-5^2=0$$ or
$$\alpha^4-14\alpha^2+9=0.$$
From here $$\alpha^6=14\alpha^4-9\alpha^2=14(14\alpha^2-9)-9\alpha^2=187\alpha^2-126,$$ which gives $$(a,b,c,d)=(-126,0,187,0).$$
