Question about simplifying $(a+b+c)(ab+bc+ac)=abc$

Question

To simplify this expression, according to math110's answer, we can write LHS as:

$$(a+b+c)(ab+bc+ac)=(a+b)(b+c)(a+c)+abc$$ then $abc$ cancels and we get: $(a+b)(b+c)(a+c)=0$

But my question is, how we can recognize that we have this equality? I have no clue about it. is there any way to quickly realize we can write $(a+b+c)(ab+bc+ac)$ as $(a+b)(b+c)(a+c)+abc$ ?

Answer 1

The original equation is

$$\frac1a + \frac1b+ \frac1c = \frac1{a+b+c}$$

We see that $a=-b, b=-c, c=-a$ are trivial solutions to the equation. Thus

$$abc(a+b+c)\left(\frac1a + \frac1b+ \frac1c - \frac1{a+b+c}\right)$$

must contain $(a+b)(b+c)(c+a)$ as a factor; indeed it is equal to $(a+b)(b+c)(c+a)$.