Better method to show $-x^3 + 1 = (-x + 1)(x^2 + x + 1)$ Is there a rule, algorithm, or theorem used for this equality below:
$$-x^3 + 1 = (-x + 1)(x^2 + x + 1)$$
I know the RHS can be established using trial and error distributing the first term through the second, or alternatively same with FOIL.  I'm curious if there is a better method?
Image below from a CAS.

 A: Let
$$S_n:=1+x+x^2+\cdots x^n.$$
Then mentally, by cancellation of the terms,
$$S_n-xS_n=1-x^{n+1},$$
and of course,
$$S_n-xS_n=(1-x)S_n.$$
A: If two $n$th degree polynomials agree on $n+1$ points, then the polynomials are equal.  You have $3$rd-degree polynomials, so choose $4$ nice values for $x$ and plug them in on both sides.  How about $x=-1, 0, 1,$ and $2$.   If you get the same value on both sides all four times, the polynomials are equal.
A: The factor theorem states that if $f(a)=0$ for some polynomial $f(x)$ then $(x-a)$ is a factor of $f(x)$. Try applying that to your case by letting $a=1$.
A: $$1-x^3=1^3-x^3$$
In general, we have$$a^n-b^n=(a-b)\sum_{k=0}^{n-1}a^{n-1-k}b^{k}$$
$a-b$ is clearly a factor of $a^n-b^n$ as when $a=b$, $a^n-b^n$ vanishes.
Another way is to view $\sum_{k=0}^{n-1}a^{n-1-k}b^{k}$ as a geometric sum.
A: Let $\zeta$ be a primitive cube root of unity, then $\zeta^3=1,\zeta^2+\zeta+1=0$ and hence $1-x^3=-(x^3-1)=-(x-1)(x-\zeta)(x-\zeta^2)=(1-x)(x^2-(\zeta+\zeta^2)x+\zeta^3)=(1-x)(x^2+x+1)$
Note: FOIL, which is the distributive property of multiplication and addition, is unavoidable in pretty much any proof. To try to avoid using FOIL in a proof would be like trying to avoid using the fact that $1+1=2\ne 0$ in a field of characteristic $\gt 2$ or $0$.
A: In the base-$10$ numeration,
$$\color{green}{9\cdot111=999}.$$
This translates
$$(10-1)(10^2+10+1)=10^3-1.$$

More generally, in the base $x=w+1$,
$$\color{green}{w\cdot111=www}$$ or $$(x-1)(x^2+x+1)=x^3-1.$$

This immediately generalizes to higher degrees.
