<s>[INST] Question about simplifying $(a+b+c)(ab+bc+ac)=abc$ 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 $(a+b)(b+c)(a+c)+abc$ ?
 [/INST] 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 $(a+b)(b+c)(c+a)$.
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