<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] Use
$$(a-b)^2+(a+b)^2=2(a^2+b^2)$$
Thus, here,
$$(a-b)^2+4=12$$
$$(a-b)^2=8$$
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