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January 31, 2015

January 31, 2015

Posted by **shivani** on Sunday, June 29, 2014 at 5:45am.

- maths (exponents -
**Steve**, Sunday, June 29, 2014 at 4:32pmWell, just using brute force, we have

b/(b+ab+1) + c/(c+bc+1) + a/(a+ac+1)

expand all that over a common denominator, and you get

(a^2b^2c + a^bc^2 + 2a^2bc + a^2b + ab^2c^2 + 2ab^2c + 2abc^2 + 6abc + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)

/

(a^2b^2c^2 + a^2b^2c + a^2bc^2 + 2a^2bc + a^2b + ab^2c^2 + 2ab^2c + 2abc^2 + 4abc + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c + 1)

now replace all the abc occurrences with just 1, and you have

(ab + ac + 2a + a^2b + bc + 2b + 2c + 6 + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)

/

(ab + ac + 2a + a^2b + bc + 2b + 2c + 6 + 2ab + ac^2 + 2ac + a + b^2c + 2bc + b + c)

where the top and bottom are clearly the same.

Now, for a trick of evaluation which avoids all that mess, I haven't come up with one yet.

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