# 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.

• $$(a+b)^2+(b+c)^2+(c+a)^2 = 4 + a^2 +b^2+c^2 = \ldots$$
– user284001
Jul 4, 2017 at 13:29
• Why would you think of $a^3+b^3+c^3-3abc$, when $(a+b+c)^2$ will lead you to the solution? Jul 4, 2017 at 13:42
• The accepted answer claims we know $a^2 + b^2 + c^2$, but this is not given. Do you have a typo'? Jul 4, 2017 at 19:51
• @DaanvdWoude : You're out of sync. lioness99a has corrected her answer to derive this result. Jul 7, 2017 at 4:01

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}

• You claim "we know the values of $a^2 + b^2 + c^2$ ...", but this value is not given. Have you skipped a step? Jul 4, 2017 at 19:52
• @EricTowers Ooops, misread the first line of the question as $a^2+b^2+c^2=1$..! Have edited my answer and corrected for that now Jul 4, 2017 at 19:58

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.

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

• This somehow feels very wrong.
– user312097
Jul 4, 2017 at 13:44
• @ A---B Please tell me where I am is wrong
– user275490
Jul 4, 2017 at 13:46
• I agree with @A---B. Think it is worth adding a bit of clarification about why you are able to do this. We're not saying you are wrong, just that it feels like a very uniniuitive way of doing it Jul 4, 2017 at 13:46
• @A---B: It feels wrong because it presumes (as does the original question, depending how one parses!) that the value of $(a+b)^{2} + (b+c)^{2} + (a+c)^{2}$ depends only on the values of $a + b + c$ and $ab + bc + ca$. If one grants the fact, Get-u's approach is about as elegant as one can want. (The other solutions establish this fact by algebra, of course.) Jul 4, 2017 at 15:23

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 $(a+b+c)^2$.

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.

• $(a+b+c)^2=a^2 + b^2 + c^2 +2ab+2bc+2ac$ is not a very clever identity.
– user312097
Jul 4, 2017 at 13:33
• @A---B But it is clever to consider the identity here. As a beginner you may not see this immediately. Jul 4, 2017 at 13:39

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 $4+a^2+b^2+c^2=4+2=6$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\i}{\mathrm{i}} \newcommand{\text}[1]{\mathrm{#1}} \newcommand{\root}[2][]{^{#2}\sqrt[#1]} \newcommand{\derivative}[3]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\abs}[1]{\left\vert\,{#1}\,\right\vert}\newcommand{\x}[0]{\times}\newcommand{\summ}[3]{\sum^{#2}_{#1}#3}\newcommand{\s}[0]{\space}$

$=\boxed{2a^2 + 2b^2 +2c^2 + 2ab + 2bc + 2ac}$

$2(ab + bc +ac) = 2$

$\boxed{2a^2 + 2b^2 +2c^2 + 2}$

$a^2 + b^2 + c^2 = (a+b+c)^2 -2ab - 2bc -2ac$

$a^2 + b^2 + c^2 = (2)^2 -2$

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

$\bbx{\color{red} {2\times 2 + 2 =6}}$

$(a+b)^2 + (b+c)^2 + (a+c)^2=a^2+b^2+c^2+2ab+2bc+2ac=2(a^2+b^2+c^2)+2(ab+bc+ac)=2(a^2+b^2+c^2)+2 \cdot 1=2((a+b+c)^2-2(ab+bc+ac))=2 \cdot (2)+2=6$

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: $$(a+b)^{2}+(b+c)^{2}+(a+c)^{2}$$ $$=2(a^2+b^2+c^2)+2(ab+bc+ca)=2\cdot 2+2=6$$