Show that $(p-1)$ is a factor of $\left(p^{10}-1\right)$ and also of $\left(p^{11}-1\right)$.

Solution:

Let $f(p)=p^{10}-1$ and $g(p)=p^{11}-1$

Putting $p=1$ in $f(p)$, we get

$f(1)=1^{10}-1=1-1=0$

Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{10}-1\right)$.

Now, putting $p=1$ in $g(p)$, we get

$g(1)=1^{11}-1=1-1=0$

Therefore, by factor theorem, $(p-1)$ is a factor of $\left(p^{11}-1\right)$.