Help proving polynomial identity

alyks
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Homework Statement



Prove the following when p is a positive integer:
b^p - a^p = (b-a)(b^{p-1}+b^{p-2}a+b^{p-3}a^2+...+ba^{p-2}+a^{p-1})

Hint: Use the telescoping property for sums.

Homework Equations


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The Attempt at a Solution



I tried reducing it to, (b-a)\sum_{k=1}^p b^{p-k}a^{k-1} but I wasn't able to do anything with it.


I've been trying to work on this exercise which is a part of a problem set in induction. But I've been having quite a bit of difficulty and I'd appreciate it if somebody here could give me a hint as to the general direction. It's been bugging me for the past few days.
 
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I don't know what telescoping is but er

(b-a)(b^{p-1}+b^{p-2}a+b^{p-3}a^2+...+ba^{p-2}+a^{p-1})

is b(b^{p-1}+b^{p-2}a+b^{p-3}a^2+...+ba^{p-2}+a^{p-1}) -a(b^{p-1}+b^{p-2}a+b^{p-3}a^2+...+ba^{p-2}+a^{p-1}).

If you can't see what that gives, expand it out by executing the multiplications and collecting up terms.

Quite a useful and important formula. When you do geometric series the same one is involved - make the relation.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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