# X^n - y^n proof

1. Sep 27, 2009

### DavidSnider

1. The problem statement, all variables and given/known data
Prove xn - yn = (x-y)(xn-1 + xn-2y + ... + xyn-2 + yn-1)

2. Relevant equations
See Above

3. The attempt at a solution
The previous problem in the book was:
Prove:
x3 - y3 = (x - y)(x2 + xy + y2)

(x - y)(x2 + xy + y2)
(x)(x2 + xy + y^2) + (-y)(x2 + xy + y2)
(x3 + x2y + xy2) + (-x2y - xy2 - y3)
x3 + x2y + xy2 - x2y - xy2 - y3
x3 - y3

I'm not sure how to show the same thing when the exponent is variable though.

2. Sep 27, 2009

### Dick

Do the same thing you did for x^3-y^3. Multiply out the right side. Many terms cancel.

3. Sep 27, 2009

### DavidSnider

I know they cancel, but.. how am I supposed to show that?

4. Sep 27, 2009

### pyrosilver

random do you happen to be using the Spivaks calculus book?

try multiplying (x-y) with all of the terms you have listedm like xn-1, xn-2y, etc and go from there

5. Sep 27, 2009

### DavidSnider

Yes, I am using Spivak's calculus. It's unlike any math book I've ever used before, so I am kind of confused as to what they are expecting me to do.

I'll think about what you just said.

6. Sep 27, 2009

### pyrosilver

I definitely agree with you. I too am using Spivak's calculus book (my class just finished chapter 2, I'm a sophomore so I'm going a little slower through the book). But yeah start by multiplying the beginning terms you have, and the end terms you have. good luck!

7. Sep 27, 2009

### Dick

Write out x*(x^(n-1)+x^(n-2)*y+...+x*y^(n-2)+y^(n-1)) and y*(x^(n-1)+x^(n-2)*y+...+x*y^(n-2)+y^(n-1)) and look for things that cancel. E.g. x*x^(n-2)*y cancels y*x^(n-1), x*x^(n-3)*y^2 cancels y*x^(n-2)*y. I know you can't write out all of the terms. You'll have to use the '...' to express what you mean. It might help to write the two expanded products on separate lines and shift one over so cancelling terms are above each other.

Last edited: Sep 27, 2009
8. Sep 27, 2009

### DavidSnider

$$(x-y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$$

$$(x)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1}) + (-y)(x^{n-1} + x^{n-2}y + ... + xy^{n-2} + y^{n-1})$$

$$(x^{n-1+1} + x^{n-2+1}y + ... + x^{2}y^{n-2} + xy^{n-1}) + (-x^{n-1}y - x^{n-2}y^{2} - ... - xy^{n-2+1} - y^{n-1+1})$$

$$x^{n} + x^{n-1}y + ... + x^{2}y^{n-2} + xy^{n-1} -x^{n-1}y - x^{n-2}y^{2} - ... - xy^{n-1} - y^{n}$$

$$x^{n} + ... + x^{2}y^{n-2} - x^{n-2}y^{2} - ... - y^{n}$$

... now I'm stuck.

Last edited: Sep 28, 2009
9. Sep 28, 2009

### Mentallic

Well, to prove $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+....+xy^{n-2}+y^{n-1})$$
we are just going to expand the RHS.

$$RHS=x(x^{n-1}+x^{n-2}y+....+xy^{n-2}+y^{n-1}) - y(x^{n-1}+x^{n-2}y+....+xy^{n-2}+y^{n-1})$$

The first factor is expanded:
$$x^n+x^{n-1}y+x^{n-2}y^2+....+x^2y^{n-2}+xy^{n-1}$$

The second factor is expanded:
$$-(x^{n-1}y+x^{n-2}y^2+....+x^2y^{n-2}+xy^{n-1}+y^n)$$

Do you notice any cancelling pattern happening?

10. Sep 28, 2009

### DavidSnider

Yeah I knew they all canceled intuitively, just wasn't sure how to show it on paper.
Filling in the blanks one step further makes it more clear.

11. Sep 28, 2009

### Mentallic

Well how about making it obvious to the examiner that you realize they cancel by lining up each equal term?

i.e. after the line $$x(x^{n-1}+x^{n-2}y+....+xy^{n-2}+y^{n-1}) - y(x^{n-1}+x^{n-2}y+....+xy^{n-2}+y^{n-1})$$

Then expand the first factor on 1 line, then expand the next factor on the line underneath, but keep cancelling factors in line with each other.

$$x^n+x^{n-1}y+x^{n-2}y^2+....+x^2y^{n-2}+xy^{n-1}$$
$$.....-x^{n-1}y-x^{n-2}y^2-.... -x^2y^{n-2} - xy^{n-1} - y^n$$

get the idea?

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