Answer:
Yes.
Step-by-step explanation:
(Assume x is not 1.)
[tex]\frac{x^n-1}{x-1}[/tex] is always divisible for integers greater than 1.
Let's use synthetic division:
1 | 1x^n + 0x^(n-1) +0x^(n-2) + ....+0x^3+0x^2+0x -1
| 1 1 1 1 1 1
---------------------------------------------------------------------------------
1 1 1 1 1 1 0
We see the remainder is 0 which means that [tex](x-1)[/tex] divides [tex](x^n-1)[/tex].
The quotient is [tex]x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^2+x+1[/tex].
