Question

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` ?

Answer 1

`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:

`1/(x^b+x^(-c)+1) + 1/(x^c+x^-a+1) + 1/(x^a+x^(-b)+1)`

`=1/3+1/3+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 = 1/(x^k+x^(-k)+1) + 1/(x^k+x^-k+1) + 1/(x^k+x^(-k)+1)`

`=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 = (x^k)^2-2(x^k)+1 = (x^k-1)^2`

So `x^k = 1`

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

Answer 2

`a+b+c=0`

Explanation:

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

`1/(x^b + x^-c + 1) + 1/(x^c + x^-a + 1) + 1/(x^a + x^-b + 1) =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 (a + b + c))` so if `n = d`

`-1 + 2 x^(a + b + c) - x^(2 (a + b + c))=0`

Solving for `x^(a + b + c)` we obtain

`x^(a + b + c) = 1` then `a+b+c=0`