Calculate the value of $ abc \cdot (a + b + c) $

Question

Let $ a, b $ and $ c $ be numbers such that $$ \begin {cases} a ^ 2-ab = 1 \\ b ^ 2-bc = 1 \\ c ^ 2-ac = 1 \end {cases} $$Calculate the value of $ abc \cdot (a + b + c) $

Attempt: Can I solve the problem by symmetry?

Answer 1

Another way.

We have $$a^2b-ab^2=b,$$ $$b^2c-c^2b=c$$ and $$c^2a-a^2c=a,$$ which gives $$\sum_{cyc}(a^2b-ab^2)=a+b+c$$ or $$(a-b)(a-c)(b-c)=a+b+c.$$ In another hand, $$\prod_{cyc}(a^2-ab)=1$$ or $$abc(a-b)(b-c)(c-a)=1,$$ which gives $$abc(a+b+c)=-1.$$

Answer 2

Hint $1$: Multiply the first equation by $bc$, the second by $ca$ and the third by $ab$ and these togther ... $e_2= ?$.

Hint $2$: Multiply the $3$ equations together ... $a^2b^2c^2=(1+ab)(1+bc)(1+ca)$ etc ...