<s>[INST] Let $a+b=2 \text{ and } ab = -1.$ Determine $a^{10} + b^{10}$ 
Let $a,b \in \mathbb{R}$ for which $$a+b=2  \text{ and } ab = -1.$$ Determine $a^{10}+b^{10}.$

I tried to approach this the following way:
From $ab=-1$ we get that $a = -\frac{1}{b}$ and substituting this to $a+b=2$ yields in $b^2-2b-1=0$. This quadratic has the solutions $b=  1 \pm \sqrt{2}$
Subsituting now $b$ we have that $a= - \frac{1}{1\pm \sqrt{2}}$
I have the feeling that I'm not going in the right direction here. Is there something I'm not seeing since the quadratic formula seems to yield a bit too much work?
 [/INST] Another way:  Following your tack, if you choose $b = 1-\sqrt{2}$ then $a=-1/(1-\sqrt{2}) = 1+\sqrt{2}$ after you rationalize the denominator.  So $a$ and $b$ are conjugates.  
If you expand $(1+\sqrt{2})^{10} + (1-\sqrt{2})^{10}$ with binomial theorem, all the odd terms will cancel out.  The 10th row of Pascal's triangle is:
1 10 45 120 210 252 210 120 45 10 1
Eliminate the odd terms and double the even and you have 
$$a^{10}+b^{10} = 2\left(1 + 45\cdot 2+210\cdot 4 + 210\cdot 8 + 45\cdot 16 + 32\right)  $$
$$=6726.$$
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