Too long problem

Notice that the equation

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}$

can be rewritten as

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

We then notice that

$(a+b+c)(ab+bc+ac) - abc = (a+b)(b+c)(a+c) = 0$

Therefore, one of the three factors must be zero. Making the substitutions $a = x+y$ $b = y+z$ $c = x+z$ shows us that one of $2x+y+z$ $x+2y+z$ $x+y+2z$ must be zero. Choose one of these without loss of generality and the proof continues as per above.

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While the problem is beautiful to solve the "real way", there is a way to cheat the solution, and that is to simply find any values of x, y, and z that satisfy the first equation and plug into the second equation.

The substitution a = x+y, b = y+z, c = x+z helps make finding these values even easier.

x = 2, y = 0, z = -1

1/(x+y)+1/(y+z)=1/[2(x+y+z)]-1/(z+x)

(x+2y+z)/[(x+y)(y+z)]=-(x+2y+z)/[2(x+y+z)(z+x)]

Luego: (Hence)

x+2y+z=0

Reemplazamos en lo que se pide: (Replace in the final expression)

i) x+y+z=-y
ii) x+y=-(z+y)
iii) x+z=-2y

(Simplify to get the answer of 2)

Saludos (Cheers)

Substitute $x+y = a$ , $y+z = b$ , and $z+x = c$ .

Notice that the given information can be rewritten as $\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{abc}$ . Conveniently, this is realized when $a = -b$ . (The main motivation for this observation is that due to Titu's Lemma, $\displaystyle \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \ge \frac{9}{abc}$ . However, Titu's lemma only applies to positive reals, and this prompts the solver to force one of the variables to be negative.

Then, the expression can be rewritten as:

$\displaystyle \frac{(a+b+c)^6 -a^6 -b^6 -c^6}{(ab)^3 +(bc)^3 + (ca)^3}$

$\displaystyle = \frac{(a-a+c)^6 -a^6 -(-a)^6 -c^6}{(-a^2)^3 + (-ac)^3 + (ac)^3}$

$\displaystyle = \frac{c^6 - c^6 -a^6 -a^6}{-a^6 - (ac)^3 + (ac)^3}$

$\displaystyle = \frac{-2a^6}{-a^6}$

$\displaystyle = \boxed{2}$

This is simple,

Define

$\begin{aligned}x+y+z &= t\\ \Rightarrow \frac{1}{t - z} + \frac{1}{t-x}\nonumber \\ \qquad{} + \frac{1}{t-y } &= \frac{1}{2t} \\\\ \text{Taking}~ x = 0 \text{ makes no }&\text{harm}\\ \Rightarrow t&= y +z\\ \Rightarrow \frac{1}{y} + \frac{1}{t} + \frac{1}{t-y } &= \frac{1}{2t} \\ \Rightarrow 2t^2 &= y(y-t)\\ \Rightarrow y &= 2t \text{ and } z = -t \end{aligned}$

$\begin{aligned}\frac{64\left( x + y + z \right) ^{6}- \left( x + y\right) ^{6}- \left( y + z\right) ^{6}- \left( z + x\right) ^{6}}{\left[ \left( x + y \right) ^{3}\left( y + z\right) ^{3}\right] + \left[ \left( y + z\right) ^{3}\left( z + x\right) ^{3}\right] + \left[ \left( z + x \right) ^{3}\left( x + y\right) ^{3}\right] } \nonumber \\ \\ \qquad=\frac{64t^6 - 64t^6 - t^6 - t^6}{8t^3 \cdot t^3 - t^3 \cdot t^3 - t^3\cdot 8t^3}\nonumber \\ \qquad=\frac{2t^6}{t^6}\nonumber \\ \qquad=\boxed{2}\end{aligned}$