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1/(x^b + x^(-c) + 1) + 1/(x^c+ x^(-a)+1) + 1/ (x^a + x^(-b) + 1) = 1, find a + b + c?

Update:

@ iceman

Can you explain how you got the third step?

4 Answers

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  • iceman
    Lv 7
    9 years ago
    Favorite Answer

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

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

    [x^(a+b+c) - 1]^2/[(x^(a + b) + x^b +1) (x^(a + c) + x^a + 1) (x^(b + c) + x^c + 1)] = 0

    x^(a + b + c) = 1

    a + b + c = 0

  • ?
    Lv 7
    9 years ago

    1

  • Brenda
    Lv 7
    9 years ago

    x^0 =1

    by having a=b=c=0

    we will have:

    1/(x^0 +x^(-0)+1) +1/(x^0+x^(-0)+1) +1/(x^0+x^(-0)+1) =>

    1/3 +1/3 +1/3 =1

    which satisfies the condition

    a+b+c=0

    ---------------

    Iceman is right.

    In each denominator change a bit it's form:

    x^b+x^(-c)+1 = x^b +1/x^c +1 =>x^bx^c+x^c+1 =>

    (x^(b+c) +x^c +1)/x^c

    the same for the others:

    x^c+x^(-a)+1 = (x^(a+c) +x^a +1)/x^a

    x^a +x^(-b) +1 = (x^(a+b)+x^b +1)/x^b

    so it becomes:

    x^c/(x^(b+c)+x^c +1)+x^a/(x^(a+c)+x^a+1)+x^b/(x^(a+b)+x^b+1)=1

    after moving the 1 to the left part... It becomes an UGLY algebra

    which simplifies to only [x^(a+b+c)-1]^2

    I think there's no enough space to show all my simplification

    of the numerator...

  • 9 years ago

    0

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