Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$ Prove that if $(a + b + c)(ab + ac + bc) = abc$ then sum of some two numbers equals $0$.
Without loss of generality let's suppose that $a=0$ then $(b + c)bc = 0 \Rightarrow b+c = 0$ or
$bc = 0 \Rightarrow b=0 \lor c = 0$ and $a+b=0 \lor a+c=0$ respectively. Q.E.D.
Now let's suppose $abc \not = 0$ and $a+b+c \not = 0$ then from initial equation we can get the following one that must hold $\dfrac{1}{a+b+c} = \dfrac{1}{a} + \dfrac{1}{b} + \dfrac{1}{c}$
As $a,b,c$ are some arbitrary numbers (but not non-negative integers for example) I've got completely stuck here

Hint:
$$(a + b + c)(ab + ac + bc) - abc=(a + b)(b + c)(c + a)$$
