# Rudin Theorem 1.21. How does he get The identity ?

1. Nov 8, 2013

Rudin Theorem 1.21. How does he get "The identity"?

In Theorem 1.21, Rudin says:

The identity $$b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+....+a^{n-1})$$yields etc etc.

What is this "identity", and do we need to prove it first? If not, what assumption is Rudin making?

2. Nov 8, 2013

### lavinia

Just multiply it out

3. Nov 8, 2013

So if we have $$b^3-a^3$$, we get
$$(b.b.b)-(a.a.a)$$

I'm probably missing something obvious, but I'm not sure what to do from here on.

4. Nov 8, 2013

### Axiomer

$(b-a)(b^{n-1}+b^{n-2}a+....+a^{n-1}) = b(b^{n-1}+b^{n-2}a+....+a^{n-1})-a(b^{n-1}+b^{n-2}a+....+a^{n-1}) = (b^{n}+b^{n-1}a+....+ba^{n-1}) - (b^{n-1}a+b^{n-2}a^2+....+a^{n}) = b^n-a^n$

5. Nov 8, 2013

Thanks that works!

It's curious though. We've proved it in one direction, but I wonder how someone made the formula in the first place.

6. Nov 8, 2013

### Axiomer

I suppose one might start by noticing that $(b-a)$ factors $b^n-a^n$, since $a$ is clearly a root of the polynomial $x^n-a^n$. So $b^n-a^n=(b-a)p(a,b)$ for some polynomial $p(a,b)$. Then, to determine $p(a,b)$, one could start doing long division to see the pattern. Alternatively, we can construct $p(a,b)$ by noticing that $p(a,b)=p(b,a)$, and that $p(a,b)$ must have $b^{n-1}$ as a term.

7. Nov 9, 2013