A) All natural numbers n
B) All odd natural numbers n
C) All even natural numbers n
D) None of the above
Correct Answer: B
Solution :
Let \[(x+y)\]is a factor of \[{{x}^{n}}+{{y}^{n}}.\] Then, \[x=-y\] will satisfy\[{{x}^{n}}+{{y}^{n}}\], \[\forall \]odd\[n\in N\]. \[\forall \,\,even\,\,n\in N.\,\,{{x}^{n}}+{{y}^{n}}\ne 0\] Hence \[{{x}^{n}}+{{y}^{n}}.\] has a factor \[(x+y)\] \[\forall \]odd\[n\in N\]. Alternate method:- \[\therefore \] \[(x+y)\] is a factor of \[{{x}^{n}}+{{y}^{n}}.\] \[\Rightarrow \,\,\,\,\,\,{{x}^{n}}+{{y}^{n}}=0\Rightarrow {{x}^{n}}+{{y}^{n}}.\] \[\Rightarrow \,\,\,\,\,\,{{x}^{n}}={{\left( -y \right)}^{n}}\,\,for\,\,all\,\,odd\,\,n.\] \[\Rightarrow \,x+y=0\]
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