Question

It is given that n is an odd integer greater than 3, but n is not a multiple of 3. then show that $x^3+x^2+x$ is factor of $(x+1{)}^n-x^n-1.$

Answer 1

Since $n$ is not a multiple of $3$, but odd integer and $x^2+x^3+x=0$
$\Rightarrow x=0,w,w^2$
Now when $x=0$
$\Rightarrow {(x+1)}^n-x^n-1-0-1=0$
$\therefore x=0$ is root of ${(x+1)}^n{-x}^n-1$
Again when $x=w$
$\Rightarrow {(x+1)}^n-w^n=1={(1+w)}^n-w^n-1=-w^{2n}-w^n-1=0$
(as $b$ is not multiple of $3$)
Similarly $x=w^2$ is root of $({(x+1)}^n-x^n-1)$
Hence $x=0,w,w^2$ are the roots of ${(x+1)}^n-x^n-1$
Thus $x^3+x^2+x$ divides ${(x+1)}^n-x^n-1$