$\{a,b,c\}\subset \Bbb R$, $a^2-ab=1,b^2-bc=1, c^2-ca=1$. Find $abc(a+b+c)$.

Consider that $\{a,b,c\}\subset \Bbb R$ and it is known that $$a^2-ab=1,\ b^2-bc=1,\ c^2-ca=1\ \ \text{(cond.1)}$$ Find $abc(a+b+c)$.

This question is from OBM 2007 (Brazilian Math Olympiad). Sorry if it is a duplicate. I will present my solution below but I think there might be easier approaches.

My attempt:

By multiplying the left and right sides from all equations from (cond. 1) we get: $$abc(a-b)(b-c)(c-a)=1$$ Multiplying the 3 terms in parenthesis leads to $$abc(abc-a²b-ac²+a²c-cb²+ab²+bc²-abc) =1$$ $$abc(a²b-ac²+a²c-cb²+ab²+bc²) =1$$ $$abc(-a(c²-ac)-c(b²-bc)-b(a²-ab))=1$$ Now by replacing the result from each equation in (cond. 1), and rearranging the expression, we get the desired answer $$abc(a+b+c)=-1$$

Question: are there other ways to solve this problem? Please provide your own answer if it uses a different approach.

• It seems you are right. What is your question? Sep 16, 2017 at 21:47
• @Piquito I'm looking to different approaches to the answer to this question. I'm studying questions similar to this one and it often helps knowing different solution techniques, perhaps with simpler arguments. Sep 16, 2017 at 21:53
• I´ve just edited the statement to make this intention clearer. Sep 16, 2017 at 21:57

My approach (not necessary simpler than yours):

Let $S=abc(a+b+c)$ then: $$S=a^2bc+b^2ac+c^2ab=(1+ab)bc+(1+bc)ac+(1+ca)ab=ab+bc+ac+S$$

So: $ab+bc+ac=0$

$$a^2+b^2+c^2-(ab+bc+ca)=3$$

So $a^2+b^2+c^2=3$

Then:

$$0=(ab+bc+ac)^2=a^2b^2+b^2c^2+c^2a^2+2S=a^2(1+bc)+b^2(1+ac)+c^2(1+ab)+2S=3+3S$$

So $S=-1$

Alt. hint:  writing the equations as $\,b=a-1/a, c=b-1/b, a=c-1/c\,$ and adding together:

$$1/a+1/b+1/c=0 \;\;\iff\;\; ab+bc+ca=0 \tag{1}$$

Then, eliminating $b$ and $c$ successively gives the equation in $a\,$:

$$3 a^6 - 9 a^4 + 6 a^2 - 1 = 0 \tag{2}$$

By symmetry, $b$ and $c$ are roots of the same equation. Morevover, if $a,b,c$ satisfy the original equations, then so do $-a,-b,-c\,$, so the $6$ roots of $(2)$ are in fact $a,b,c,-a,-b,-c\,$.

It follows from Vieta's formulas that $\,-(abc)^2=-1/3\,$ and $\,-a^2-b^2-c^2=-9/3=-3\,$. Then, given $\,(1)\,$, it further follows that $\,(a+b+c)^2=a^2+b^2+c^2 = 3\,$, so $\,\big(abc(a+b+c)\big)^2=1\,$. (This still leaves the choice of sign for $\,\pm 1\,$ to figure out, of course.)