The question given is

If $\dfrac{(b−c)}{a} + \dfrac{(a+c)}{b} + \dfrac{(a−b)}{c}=1$ and $a-b+c \neq 0 $ then prove that $\dfrac 1a = \dfrac 1b + \dfrac 1c$

I tried to take $abc$ on the right hand side after taking the LCM, but ended up with $b^2(c-a)+a^2(b+c)+c^2(a-b)=abc$. I could not simplify any further. Please provide only hints, not complete solution.