about to only mathematics

Explanation

Since n is not a multiple of 3, but odd intergers and `x^3+x^2 +x= 0 rArr x+0 , omega , omega^2`
Now when x=0
`rArrr (x+1)^n-x^n-1=1-0-1=0`
`therefore x=0 is root of (x+1)^n-x^n-1`
as n is a multiple of 3 of and odd] Similarly , `x=omega^2` is root of `{(x+1)^n-x^n-1`
Hence x=0 `omega = omega^2` are the roots of `(x+1)^n-x^n-1`
Thus `x^3+x^2 +x "divides" (x+1)^n -x^n -1`