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pratu043
 2 years ago
Best ResponseYou've already chosen the best response.0How do you prove that?

Libniz
 2 years ago
Best ResponseYou've already chosen the best response.0sum of odd roots factor always factor out (x+y)

pratu043
 2 years ago
Best ResponseYou've already chosen the best response.0What do you mean by that?

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2I know how, PLZ wait sec trying to write it down

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Oh got it.... let f(x)=x^n + y^n

pratu043
 2 years ago
Best ResponseYou've already chosen the best response.0So you should use factor theorem for this?

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2now, when f(x) is divided by (x+y) then REMAINDER = f(y) = (y)^n + y^n

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Now, for x^n + y^n to be divisible by x + y.. remainder = 0 or, (y)^n + y^n=0

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2which is only possible when n is odd

pratu043
 2 years ago
Best ResponseYou've already chosen the best response.0ok so when n is odd, (y)^n = y so (y)^n + y^n = y + y = 0 thanks!!

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2oh its like this: when n is odd, (y)^n = (y)^n so (y)^n + y^n = (y)^n + y^n = 0
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