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This isn't a homework question. I am part of a math team and I am working on my corrected score of a practice test. I need to know how to do these kinds of problems to prepare for competitions. Any help you have will be appreciated.

Find the value of k for which log((x+y)/k) = log(x)/3 + log(y)/3 + log(x+y)/3 for all positive values of x and y for which x^3 + y^3 = 340xy(x+y).

Thank you

Use the identity x^3 + y^3 = (x+y)^3 - 3xy(x+y)

x^3 + y^3 = 340xy(x+y)

(x+y)^3 - 3xy(x+y) = 340xy(x+y)

(x+y)^3 = 343xy(x+y)

(x+y)^3 = (7^3)xy(x+y)

(x+y)^3 / (7^3) = xy(x+y)

[(x+y)/7]^3 = xy(x+y)

log [(x+y)/7]^3 = log xy(x+y)

3 log [(x+y)/7] = log(x) + log(y) + log(x+y)

log [(x+y)/7] = (1/3)log(x) + (1/3)log(y) + (1/3)log(x+y)

so k = 7

Problem solving is a huge topic , there are many sites that offer tips and advice.

As we see in this problem , it's often a good idea to express a symmetric function in terms of the elementary symmetric functions. Also , remember the rules for logarithms.

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