<s>[INST] Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 −  + 1$;  = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 −  + 1$;  = 1, 2, 3,...

I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$;  = 1, 2, 3,..
Basic Step
Let n = 1
⇒ $x^2 - x + 1$ | $^8 − ^7 + 1$
I then proved that the remainder is 0 using polynomial long division.
$\frac{^{6n+2} − ^{6n+1} + 1}{x^2 - x + 1}$ = $x^6 - x^4 - x^3 + x + 1$ R 0
∴ s(1) is true
Assumption Step
Assume that s(m) is true
⇒ $\frac{^{6m+2} − ^{6m+1} + 1}{x^2 - x + 1}$ = Q(x) where Q(x) is a polynomial
Inductive Step
To prove that s(m+1) is true
⇒ $\frac{^{6(m+1)+2} − ^{6(m+1)+1} + 1}{x^2 - x + 1}$ = T(x) where T(x) is a polynomial(x) is a polynomial
⇒ $\frac{^{6m+8} − ^{6m+7} + 1}{x^2 - x + 1}$ = T(x)
⇒ $\frac{x^6(^{6m+2} − ^{6m+1}) + 1}{x^2 - x + 1}$ = T(x)

However, I'm unsure of how to proceed from here. I would appreciate it if anyone could help me with this. Thanks!
 [/INST] *

*Finds the roots of $x^2-x+1$; they are complex

*Pick one of them and prove that is a root of $x^{6n+2}-x^{6n+1}+1$; automatically the other one is also a root therefore the two polynomials have common roots, hence the one with lesser number of roots is a factor of the other polynomial.

</s>