<s>[INST] If $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$, for integers $a$ and $b$, then what is $\left(1-\sqrt2\right)^{2010}$ expressed using $a$ and $b$? 
Being $ a $ and $ b $ integers such that $\left(1+\sqrt{2}\right)^{2011} =a+b\sqrt{2}, \left(1-\sqrt{2}\right)^{2010}$ equals:
a) $a+2b+(a-b)\sqrt{2}$
b) $a-2b+(a-b)\sqrt{2}$
c) $a+2b+(b-a)\sqrt{2}$
d) $2b-a+(b-a)\sqrt{2}$
e) $a+2b-(a+b)\sqrt{2}$

Solution:

Not going with any of the alternatives
 [/INST] The key point is that $\left(1+\sqrt2\right)^{2011}=a+b\sqrt{2}$ implies
$\left(1-\sqrt2\right)^{2011}=a-b\sqrt{2}$.
Therefore,
$$
\left(1-\sqrt2\right)^{2010}
=\frac{\left(1-\sqrt2\right)^{2011}}{1-\sqrt2}
=\frac{\left(1-\sqrt2\right)^{2011}(1+\sqrt2)}{(1-\sqrt2)(1+\sqrt2)}
=-(a-b\sqrt2)(1+\sqrt2)
= \cdots
$$
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