# If 1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1) = 1 then what can we say about a, b, c ?

Aug 21, 2016

$a + b + c = 0$

#### Explanation:

For any non-zero value of $x$, we have ${x}^{0} = 1$.

So with $a = b = c = 0$ we have:

$\frac{1}{{x}^{b} + {x}^{- c} + 1} + \frac{1}{{x}^{c} + {x}^{-} a + 1} + \frac{1}{{x}^{a} + {x}^{- b} + 1}$

$= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

Actually as seen in https://socratic.org/s/axdYQgwe, if $a + b + c = 0$ then this equation holds for any non-zero value of $x$.

Note also that if $a = b = c = k$ then:

$1 = \frac{1}{{x}^{k} + {x}^{- k} + 1} + \frac{1}{{x}^{k} + {x}^{-} k + 1} + \frac{1}{{x}^{k} + {x}^{- k} + 1}$

$= \frac{3}{{x}^{k} + {x}^{- k} + 1}$

So we have:

${x}^{k} + {x}^{- k} + 1 = 3$

Subtracting $3$ from both sides and multiplying through by ${x}^{k}$ we get:

$0 = {\left({x}^{k}\right)}^{2} - 2 \left({x}^{k}\right) + 1 = {\left({x}^{k} - 1\right)}^{2}$

So ${x}^{k} = 1$

This is satisfied for any non-zero value of $x$ if $k = 0$ and no other values of $k$.

Aug 21, 2016

$a + b + c = 0$

#### Explanation:

Using "brute force" or with the help of a symbolic processor,

$\frac{1}{{x}^{b} + {x}^{-} c + 1} + \frac{1}{{x}^{c} + {x}^{-} a + 1} + \frac{1}{{x}^{a} + {x}^{-} b + 1} = \frac{n}{d} = 1$

n = (x^a + x^b + 2 x^(a + b) + x^(2 a + b) + x^c + 2 x^(a + c) + 2 x^(b + c) + 6 x^(a + b + c) + 2 x^(2 a + b + c) + x^(2 b + c) + 2 x^(a + 2 b + c) + x^(2 a + 2 b + c) + x^(a + 2 c) + 2 x^(a + b + 2 c) + x^(2 a + b + 2 c) + x^(a + 2 b + 2 c))

d = (1 + x^a + x^b + 2 x^(a + b) + x^(2 a + b) + x^c + 2 x^(a + c) + 2 x^(b + c) + 4 x^(a + b + c) + 2 x^(2 a + b + c) + x^(2 b + c) + 2 x^(a + 2 b + c) + x^(2 a + 2 b + c) + x^(a + 2 c) + 2 x^(a + b + 2 c) + x^(2 a + b + 2 c) + x^(a + 2 b + 2 c) + x^( 2 a + 2 b + 2 c))

but

$n - d = - 1 + 2 {x}^{a + b + c} - {x}^{2 \left(a + b + c\right)}$ so if $n = d$

$- 1 + 2 {x}^{a + b + c} - {x}^{2 \left(a + b + c\right)} = 0$

Solving for ${x}^{a + b + c}$ we obtain

${x}^{a + b + c} = 1$ then $a + b + c = 0$