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.


$$a+b=2$$
$$\implies(a+b)^2=4$$
$$\implies a^2+b^2+2ab=4$$
$$\implies2ab=-2$$

Also,

$$(a-b)^2=a^2+b^2-2ab$$
$$6-(-2)=6+2=8$$(by substituting $a^2+b^2=6$ and $2ab=-2$)

