<s>[INST] Find the value of $(a - b)^2$ given $a + b = 2$ and $a^2 + b^2 = 6$ 
$$a + b = 2\\
a^2 + b^2 = 6$$ 
Find the value of $(a-b)^2 $

My workings till I got stuck -
$$(a-b)^2 = a^2 - 2ab + b^2 \\
= a^2 + b^2 - 2ab\\ 
= 6 - 2 ab $$
I'm stuck at how to find $ab$ . Can I get hints on how to find $ab$? Thanks a lot.
 [/INST] We will solve for the values of $a$ and $b$. $a=2-b$ so $a^2+b^2=6 \iff (2-b)^2 + b^2 +6 $
This means $2b^2-4b-2=0 \iff b^2 -2b -1 +0 $. The solutions of this equation are $b=1-\sqrt{2}$ and $b=1+\sqrt{2}$. Then $a=1+\sqrt{2}$ or $a=1-\sqrt{2}$
Then, in any case $(a-b)^2=8$
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