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. 
 A: 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}
A: 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.
$$(1)^2 + (2)^2 + (1)^2 = 6$$
A: 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 $(a+b+c)^2$.
A: Either you are using a clever identity (see the other answers), or you just substitute $c=2-a-b$ so that
$$
ab+bc+ca-1=- a^2 - ab + 2a - b^2 + 2b - 1,
$$
and compare it to the identity in question
$$
(a+b)^2+(b+c)^2+(c+a)^2=2a^2 + 2ab - 4a + 2b^2 - 4b + 8.
$$
This may not be elegant, but at least you know how to do it yourself.
A: 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 $4+a^2+b^2+c^2=4+2=6$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}$
$=\boxed{2a^2 + 2b^2 +2c^2 + 2ab + 2bc + 2ac}$
$2(ab + bc +ac) = 2$
$\boxed{2a^2 + 2b^2 +2c^2 + 2}$
$a^2 + b^2 + c^2 = (a+b+c)^2 -2ab - 2bc -2ac$
$a^2 + b^2 + c^2 = (2)^2 -2$
$a^2 + b^2 + c^2 = 2$
$\bbx{\color{red} {2\times 2 + 2 =6}}$
A: $(a+b)^2 + (b+c)^2 + (a+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=2(a^2+b^2+c^2)+2(ab+bc+ac)=2(a^2+b^2+c^2)+2 \cdot 1=2((a+b+c)^2-2(ab+bc+ac))=2  \cdot (2)+2=6$
A: Consider the monic polynomial with roots $a$, $b$, $c$
$$P(x)=(x-a)(x-b)(x-c)$$
Expanding out:
$$P(x)=x^{3}+x^{2}(-a-b-c)+x(ab+bc+ac)-abc$$
(This is essentially Vieta's formula).
Then plugging in your values this polynomial equals:
$$P(x)=x^{3}-2x^{2}+x-abc$$
Further, since $a$, $b$, $c$ are roots
$$P(a)=a(a^{2}-2a+1-bc)=0$$
$$P(b)=b(b^{2}-2b+1-ac)=0$$
$$P(c)=c(c^{2}-2c+1-ab)=0$$
We assume $a,b,c$ are non-zero. Then division gives us:
$$P(a)=a^{2}-2a+1-bc=0$$
$$P(b)=b^{2}-2b+1-ac=0$$
$$P(c)=c^{2}-2c+1-ab=0$$
Adding up the above three, and using $a+b+c=2$ and $ab+bc+ac=1$
$$a^{2}+b^{2}+c^{2}=2(a+b+c)-3+(ab+bc+ca)$$
$$=2$$
Now we can tackle your problem. Just expand:
$$(a+b)^{2}+(b+c)^{2}+(a+c)^{2}$$
$$=2(a^2+b^2+c^2)+2(ab+bc+ca)=2\cdot 2+2=6$$
