If the roots of the equation $x^n-1=0$ are $1,\alpha,\beta,\gamma,\cdots$, prove that $(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$

My first post here

So I was doing some past year questions and this one popped out and I haven't been able to progress much upon it and I believe there's a trick which will get it done in no time.

If the roots of the equation $$x^n-1=0$$ are $$1,\alpha,\beta,\gamma,\cdots$$

we need to show that $$(1-\alpha)(1-\beta)(1-\gamma)\cdots= n$$

Can I be helped with some hints to push me in the right direction to solve it?

Apparently

$$x^n-1 = (x-1)(x-\alpha)(x-\beta)(x-\gamma)...$$

setting $$P(x) = \tfrac{x^n-1}{x-1}$$

$$P(x)= (x-\alpha)(x-\beta)(x-\gamma)...$$

Now just evaluate at x = 1.

• This seems to be the 'trick' i was exactly looking for, thanks Mar 11, 2021 at 7:56
• @Arthur Yes, it works out fine though, since $P(x)$ has a closed polynomial form. Mar 11, 2021 at 8:01
• @Arthur that does not require limits: $(x^n-1)/(x-1) = x^{n-1} + x^{n-2} + \cdots + x + 1$ as polynomials and the value of the right side at $x=1$ is $n$. Polynomials equal at infinitely many numbers are equal everywhere because a nonzero polynomial has only finitely many roots.
– KCd
Mar 11, 2021 at 8:01
• @KCd You're right. I missed that one. Mar 11, 2021 at 8:22

Denote the roots by $$x_0 = 1, x_1, x_2, x_3, \dots, x_{n-1}$$
(roots of unity, usual indexing in counter-clockwise direction)

Then $$x_k = x_1^k$$

So this whole product is an expression involving just $$x_1$$

And from the roots of unity theory we know that

$$x_1 = \cos\frac{2\pi}{n} + i \cdot \sin\frac{2\pi}{n}$$

Proceeding from here is a matter of doing some simple algebra.

• $x_k = x_1^k$ really depends on which of the roots you call $\alpha$. It isn't necessarily the first counterclockwise root after $1$. I mean, having $\alpha$ be any other root would be confusing and obfuscating. But it should be clarified. Mar 11, 2021 at 7:54
• @Arthur Yes. I call $x_1$ the usual one, the one from my formula. Mar 11, 2021 at 7:55
• @peter.petrov My point is, you don't know that that's $\alpha$. And you say $x_1=\alpha$. It's a little pedantic, but I think it ought to be clarified. Mar 11, 2021 at 7:56
• @Arthur Yeah, it's pedantic. I removed any references to $\alpha$ anyway. Mar 11, 2021 at 8:04

using Transformation of equations, let $$1-x=y$$

$$\implies(1-y)^n=1\iff y^n-\binom n1y^{n-1}+\cdots+\binom n{n-1}(-1)^{n-1}y=0$$ whose roots are $$y_k, 0\le k\le n-1$$

Clearly, one of the them is $$0,$$ let $$y_0=0$$

So, the roots of $$y^{n-1}-\binom n1y^{n-2}+\cdots+\binom n{n-1}(-1)^{n-1}=0$$ will be $$y_k, 1\le k\le n-1$$

Finally use Vieta's formula

If the roots are $$r_i$$, $$i=1\ldots n$$, with $$r_1=1$$, we have

$$\prod\limits_{i=1}^{n}r_i= -(-1)^n$$

$$\sum\limits_{i=1}^{n}r_i = \sum\limits_{i (till taking $$n-1$$ terms at a time)

Now, $$\sum\limits_{i=2}^{n}r_i=\sum\limits_{i=1}^{n}r_i-1=0-1=-1$$

Also, $$\sum\limits_{i,j=2,i

We can show that the terms $$\sum\limits_{i=2}^{n}r_i, \sum\limits_{i,j=2,i have alternating values $$-1$$ and $$1$$

$$\therefore (1-\alpha)(1-\beta)(1-\gamma)\ldots$$

$$=\prod\limits_{i=2}^{n}(1-r_i)$$

$$=1-\sum\limits_{i=2}^{n}r_i+\sum\limits_{i,j=2}^{n}r_ir_j-\ldots$$

$$=1-(-1)+1-(-1)+\ldots=n$$