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In algebra & polynomials, how do we prove that

$$x-1 \mid x^n -1?$$

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Consider $(x-1)(1+x+x^2+\ldots+x^{n-1})$.

You can prove it by induction. $$x^{n+1}-1 = x^{n+1}-x^n + x^n -1 = (x-1)x^n + x^n-1$$

Clearly $x\equiv 1 \pmod{x-1}$. Hence $x^n-1\equiv 1^n-1\equiv 0 \pmod{x-1}$, i.e. $x-1\mid x^n-1$.

Hints: For any two reals $\,a,b\,$ :

$$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\ldots+ab^{n-2}+b^{n-1})$$

**Factorization method**

Clearly, $x^{n} – 1 = (x – 1)(x^{n – 1} + x^{n – 2} + \dots + 1)$. Thus, $(x – 1) | (x^{n} – 1)$ since there exists the polynomial namely $(x^{n – 1} + x^{n – 2} + \dots + 1)$ that is multiplied by $(x – 1)$ to obtain $(x^{n} – 1)$.

**Substitution method**

Suppose that $x – 1 = 0 \rightarrow x = 1$. If we substitute that value for $(x^{n} – 1)$, then clearly $1^{n} – 1 = 0$. This shows that $(x^{n} – 1)$ has a factor $(x – 1)$.

Hint: Apply division algorithm, what can be the remainder?

By Remainder theorem $ f(x) = x^n -1 ,$ if $ f(1) = 0$, then $ (x-1)$ is a factor of $f(x) $ .

Another proof by induction:

$$

x^{n+1}-x+x-1=x(x^{n}-1)+x-1

$$

RHS also divides $x-1$ because $x^n-1$ divides it by the assumption of the $n^{th}$ step

$x-1$ has as many zeros as its degree allows (i.e. one, and it’s 1), and $x^n-1$ shares all those zeros. From that it follows that $x-1|x^n-1$, at least of you’re working over a field. Over a ring (i.e. where you cannot, in general, divide) you also need that the coefficient of the largest power in $x-1$ divides the coefficient of the largest power in $x^n-1$, but you have that too.

What you use here is that if $p(a)=0$, then $(x-a)|p$.

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