Question

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

Answer 1

we have $x^3+x^2+x=x[x^2+x+1]=x(x-\omega )(x-\omega )$

$letf\left(x\right)=(x+1{)}^n-x^n-1$

in order to show that $x^3+x^2+x$ is the factor of f(x) we must show that f(x) = 0 when x = 0 , x = $\omega andx={\omega }^2$

now f(0) = $(0+1{)}^n-0-1=1-1=0$

$f\left(\omega \right)=(\omega +1)-{\omega }^n-1=(-{\omega }^2{)}^n-1$

$=(-1^n){\omega }^{2}n+{\omega }^n-1={\omega }^{2}n+{\omega }^n-1$ ,

$(-1{)}^n=-1$ since n is a odd integer

$=-[1+{\omega }^{2}n+{\omega }^n-1=0]$

similarly $f\left({\omega }^2\right)=({\omega }^2+1)-({\omega }^2{)}^2-1)$

$=(\omega {)}^n-{\omega }^{2}n-1)$

$=(\omega {)}^n-{\omega }^{2}n-1=0$

Hence $x(x-\omega )(x-\omega )thatisx{}^3+x^2+x$ is a factor of $f\left(x\right)$