$\{a,b,c\}\subset \Bbb R$, $a^2-ab=1,b^2-bc=1, c^2-ca=1$. 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.
 A: 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: $ab+bc+ac=0$
Now add the three constraints
$$a^2+b^2+c^2-(ab+bc+ca)=3$$
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 $S=-1$
A: 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.)
