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pratu043
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Find the condition that x^n + y^n may be divisible by x + y.
 2 years ago
 2 years ago
pratu043 Group Title
Find the condition that x^n + y^n may be divisible by x + y.
 2 years ago
 2 years ago

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sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
IF n is odd
 2 years ago

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

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

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

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

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

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

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

pratu043 Group TitleBest ResponseYou've already chosen the best response.0
ok got that.
 2 years ago

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

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

sauravshakya Group TitleBest ResponseYou've already chosen the best response.2
got it?
 2 years ago

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

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