Question

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

Answer 1

Since, $n$ is not a multiple of $3$, but odd integers and
$x^3+x^2+x=0 \, \Rightarrow \, \, x=0, \omega, \omega^2$
Now, when $x = 0 \Rightarrow ( x + 1)^n - x ^n - 1 = 1 - 0 - 1 = 0$
$\therefore x = 0$ is root of $(x + 1)^n- x^n- 1$
Again, when $x = \omega$
$\Rightarrow \, \, (x+1)^n-x^n-1=(1+\omega)^n-\omega^n -1$
$ =-\omega^{2n}-\omega^n-1=0$
[as $n$ is not a multiple of $3$ 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 $(x + 1)^n - x^n - 1$ .