Question

prove that$(x+y)n-x^n-y^n$ it divisible for$xy(x+y)(x^2+xy+y^2)$if n is odd but not multiple of 3

Answer 1

we have $xy(x+y)(x^2+xy+y^2)$
=$xy(x+Y)(y-x\omega )(y-x{\omega }^2)$
let$(x+y)n-x^n-y^n$ it divisible for$xy(x+y)(x^2+xy+y^2)$
considering the given expression f (x , y) as a polynomial in y , we put y =o
we see that at y = 0 the polynomial f(x , y) becomes $x^n-x^n=0$
it becomes 0 for any x . Thus f(x , y) is divided by y.
similary f(x , y) is divide by x . Thus f(x , y) is divided by xy.
To prove that f(x , y) is divided by x + y becomes$x \, - \, x)^2 \, - \, x^2 \, - \, (-x)^2
= \, 0 \, - \, x^n \, + \, x^n \, = \, 0
[ n is odd , We have (-x^2) \, = \, - x^n $
consequently our polynomial is divided by x + y It remains to prove that f(x , y) is divided by
$y-x\omega and,y=x{\omega }^2$
In f(x , y) , we put $y=\omega$ It becomes
$(x \, + \, x\omega )^n \, - \, x^n \, - \, (x\omega )^n \, = \, x^n (-\omega ^2)^n \, - \, x^n \, - \, x^n \omega ^n \, = \, x^n \, [(-1)^n \omega^{2n} \, - \, \omega^n \, - \, 1]
= \, - \, x^n [ \omega^{2n} \, - \, \omega^n \, - \, 1]
= x^n \, \times \, 0 = \, 0 $by
similary f(x , y) vanished when $y={\omega }^{n}x$
Hence is polynomial f(x , y) is divided by
$xy(x+y)(x^2+xy+y^2)$