let a be a real numer and n a positive integer.show that x-a is a factor of Xn (n square) -An (upperscript too)
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Answer:
Let f(x)=x
n
+a
n
.
In order to prove that x+a is a factor of f(x) for any odd positive integer n, it is sufficient to show that f(−a)=0.
f(−a)=(−a)
n
+a
n
=(−1)
n
a
n
+a
n
f(−a)=(−1+1)a
n
[ n is odd positive integer ]
f(−a)=0×a
n
=0
Hence, x+a is a factor of x
n
+a
n
, when n is an odd positive integer.
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