1

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 $(a+b)(b+c)(a+c)+abc$ ?

Etemon
  • 6,437

1 Answers1

3

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 $(a+b)(b+c)(c+a)$.

player3236
  • 16,413
  • From the trivial solutions I can conclude $\frac1a+\frac1b+\frac1c-\frac{1}{a+b+c}$ has $(a+b)(a+c)(b+c)$ as a factor. then you multiplied expression by $abc(a+b+c)$ to simplify the fractions. did I understand it correctly? – Etemon Feb 09 '21 at 04:01
  • We can't say that $\frac1a +\dots$ has that factor since it is not a polynomial. I multiplied $abc(a+b+c)$ to the fractions so it would become one, and $(a+b)(a+c)(b+c)$ would be a factor to that polynomial. – player3236 Feb 09 '21 at 04:05
  • Oh I got it. thank you very much for clearing my doubts! – Etemon Feb 09 '21 at 04:06