Solve $(1+c-b)(1+a-c)(1+c-a)=(1+b-c)(1+b-a)(1+a-b)$

Question

I am interested in finding real value solutions to: $(1+c-b)(1+a-c)(1+c-a)=(1+b-c)(1+b-a)(1+a-b)$

Clearly, this is trivially true when $a=b=c$. I'm wondering if there are non-trivial real solutions. There are $3$ unknowns, only one equation. I suspect there are many non-trivial solutions.

Answer 1

Another possibility

$$ (1 + c - b) (1 + a - c) (1 + c - a) - (1 + b - c) (1 + b - a) (1 + a - b) = (b-c) \left(2 a^2-2 a (b+c+1)+b^2+b+c^2+c-2\right) = 0 $$

so

$$ b = c\\ 2 a^2-2 a (b+c+1)+b^2+b+c^2+c-2 = 0 $$

which gives

$$ a = \frac{1}{2} \left(1+b+c\pm \sqrt{5-(b-c)^2}\right) $$

so the solutions appear to be also as

$$ b = c\\ a = \frac{1}{2} \left(1+b+c\pm \sqrt{5-(b-c)^2}\right) $$

Answer 2

Since $a,b,c$ are arbitrary, $x=a-b$, $y=b-c$, $z=c-a$ are also arbitrary. Hence, the equation can be rewritten as $$(1-y)\cdot(1-z^2) = (1+y)\cdot(1-x^2)$$So, $x,y,z=0$ is an obvious solution, as is $(1,1,z)$ for any $z$, or $(x,-1,1)$ for any $x$.

When $|y|\neq1$, our solution set for arbitrary $x,y$ is $$z = \pm\sqrt{\frac{x^2(1-y)+2y}{1+y}}$$Is it pretty? No, but it is your solution set.