Algebraic identities

We can expand $$(a+b)^2+(b+c)^2+(c+a)^2$$ to get

$$2 a^2 + 2 a b + 2 a c + 2 b^2 + 2 b c + 2 c^2 $$

We can rearrange this to be more useful:

\begin{align}2 a^2 + 2 a b + 2 a c + 2 b^2 + 2 b c + 2 c^2 &= 2 a^2 + 2b^2+2c^2 + 2 a b + 2 a c + 2 b c\\ &=2(a^2+b^2+c^2)+2(ab+ac+bc)\end{align}

We know the value of $ab+ac+bc$, so we can say that \begin{align}2(a^2+b^2+c^2)+2(ab+ac+bc)&=2(a^2+b^2+c^2)+2\end{align}

Now we need to find the value of $a^2+b^2+c^2$.

We can do this as follows:

\begin{align}(a+b+c)^2&=2^2\\ a^2 + 2 a b + 2 a c + b^2 + 2 b c + c^2&=4\\ a^2+b^2+c^2+2(ab+ac+bc)&=4\\ a^2+b^2+c^2+2\times 1&=4\\ a^2+b^2+c^2&=2\end{align}

So now we can say that \begin{align}2(a^2+b^2+c^2)+2&=2\times 2+2\\ &=6\end{align}

If you just want the value of expression $(a+b)^2 + (b+c)^2 + (a+c)^2$than set the vale of $a=0$, $b=1$ and $c=1$ (these satisfy the required conditions ) and you are going to get the answer.

$$(1)^2 + (2)^2 + (1)^2 = 6$$

Hint:

Express $S_2=a^2+b^2+c^2$ in function of $s=a+b+c$ and $\sigma=ab+bc+ca$ from the algebraic identity for $(a+b+c)^2$.