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!
 A: Hint
$a^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\Rightarrow a^2-7=2\sqrt{10} \Rightarrow (a^2-7)^2=40$
Edit
$(a^2-7)^2=40 \Rightarrow a^4-14a^2+49=40 \Rightarrow a^4-14a^2+9=0$
A: I think you're looking at this from the wrong perspective. Some basic Galois Theory will tell you what the other roots of the polynomial are (hint: consider how it's constructed. Whenever we conduct a squaring operation, we 'spawn' another solution corresponding to the negative of whatever quantity is being squared). But those other roots aren't actually relevant to solving the question of how to write $\alpha^6$.
Instead, consider doing some 'modular' algebra: you know that $\alpha^4=14\alpha^2-9$ (as other answers have showed). You can use this to 'reduce' any term of size $\alpha^4$ or larger in a polynomial expression to a term that's of smaller degree. For instance, by multiplying by $\alpha$ we see that $\alpha^5=14\alpha^3-9\alpha$. Now, multiply both sides of this by $\alpha$; can you see a spot to use the relation again?
A: 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 $$(a,b,c,d)=(-126,0,187,0).$$
A: We have
$$\left( x-\sqrt{3}-\sqrt{2}\right) \, \left( x-\sqrt{3}+\sqrt{2}\right) \, \left( x+\sqrt{3}-\sqrt{2}\right) \, \left( x+\sqrt{3}+\sqrt{2}\right) ={{x}^{4}}-10 {{x}^{2}}+1 \tag{1}$$
and further
$$ x^6=({{x}^{2}}+10)({{x}^{4}}-10 {{x}^{2}}+1)+(99 {{x}^{2}}-10)$$
So
$$\alpha^6=99 \alpha^2-10$$
Instead of guessing $(1)$ you can find a polynomial of degree 4 by using the method described here by @HagenvonEitzen.
