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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?
@ iceman
Can you explain how you got the third step?
4 Answers
- icemanLv 79 years agoFavourite 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
- BrendaLv 79 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
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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...
