✔ 最佳答案
Let f(x) = x^( 2n-1 ) + y^( 2n-1 )
By Factor Theorem, if f( -y ) = 0, f(x) is divisible by ( x + y ).
http://en.wikipedia.org/wiki/Factor_theorem
( A deeper approach to Factor Theorem )
Put x = -y into x^( 2n-1 ) + y^( 2n-1 )
f(x) = ( -y )^( 2n - 1 ) + y^( 2n - 1 )
No matter n is even or odd, 2n is an even number
So, ( 2n - 1 ) must be an odd number
So f(x) = - y^( 2n - 1 ) + y^( 2n - 1 )
= 0
As f( -y) = 0, f(x) is divisible by ( x + y )