Algebraic identities Given that $$a+b+c=2$$ and $$ab+bc+ca=1$$
Then the value of 
$$(a+b)^2+(b+c)^2+(c+a)^2$$ is how much?

Attempt:
Tried expanding the expression. Thought the expanded version would contain a term from the expression of $a^3+b^3+c^3-3abc$, but its not the case. 

If you expand you find its $a^2+b^2+c^2+2ab+2bc+2xa+a^2+b^2+c^2$. Now square first equation and we have $a^2+b^2+c^2+2ab+2bc+2ac=4. ..(3)$ thus now we want value of $4+a^2+b^2+c^2$ .rearranging $3$ and using your equation $2$ we have $a^2+b^2+c^2+2 (ab+bc+ac)=4$ thus $a^2+b^2+c^2+2=4$ thus value of expression we want is $4+a^2+b^2+c^2=4+2=6$
