# Spivak: Is his how to approach these proofs?

1. Sep 1, 2011

Hello all

I have started the problem set for Chapter one (basic properties of numbers) in Spivak's Calculus (self study). I think I am doing these right, but I have some questions.

As a solid example, problem 1-(iv) says to prove the following:

$$x^3 - y^3 = (x-y)(x^2+xy+y^2)\qquad(1)$$

My approach was to add and subtract terms to the left hand side of (1) until it could be factored into the desired form:

$$\begin{array}{l} x^3 - y^3 &= x^3 - y^3 + (x^2y - x^2y) + (xy^2 - xy^2) \\ &= x^3 +x^2y+xy^2 - y^3 - x^2y - xy^2 \\ &=x(x^2+xy+y^2)-y(x^2+xy+y^2) \\ &= (x-y)(x^2+xy+y^2) \end{array}$$

Now this seems correct to me, but I feel a little guilty because I only knew what to add and subtract to the LHS of (1) because I knew what I was trying to achieve- the RHS of (1).

My second question is similar, but requires some clarifying of Spivak's intent. In problem 1-(vi) he says: Prove the following:

$$x^3 + y^3 = (x+y)(x^2-xy+y^2)\qquad(2)$$

Then he says, "There is a particularly easy way to do this, using (iv), and it will show you how to find a factorization of $x^n+y^n$ whenever n is odd."

Well, I solved this one the exact same way I did with (iv) above: I used the right hand side of (2) to infer what terms to add/subtract and obtained the solution. However, the fact that I do not see what he means by the quoted text above leads me to belive that there was some other approach. That is, I have not discovered a way to factorize $x^n+y^n$ whenever n is odd in my procedure.

Any thoughts are appreciated

2. Sep 1, 2011

### LCKurtz

Certainly, if you know (1), you can get (2) by simply substituting -y in for the y.

3. Sep 1, 2011

Hi LC Yes. I arrived at the solution, but in particular, I am concerned about my approach as I do not see what he means by

4. Sep 1, 2011

### LCKurtz

OK, I see what you mean now. Apparently you need a more clever approach to (1) that easily generalizes to higher odd exponents. I dunno -- I'll think about it if this football game doesn't keep me too distracted. But don't count on me for this one

5. Sep 1, 2011

### LCKurtz

Also, I guess that observing that y = x is a root for odd exponents and factoring (y-x) by long division doesn't count?

6. Sep 1, 2011

Maybe he made a typo? He says "this can easily be done using (iv)". Maybe he meant (v). In problem (v) we proved that

$$x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y+\dots+xy^{n-2} +y^{n-1} )$$

Anyway, no big deal! Get back to your game!

7. Sep 2, 2011

### micromass

Staff Emeritus
My thoughts:

Spivak is correct, (vi) can be easily done by (iv) by just substituting -y for y.
This gives you a hint how to obtain the factorization of $x^n+y^n$. Indeed, take (v) and substitute -y for y again...

8. Sep 2, 2011

### 206PiruBlood

I am pretty sure this is what Spivak intended.

Also note that you don't have to start from the left side for the first one..

(x-y)($x^{2}$+xy+$y^{2}$)=x($x^{2}$+xy+$y^{2}$)-y($x^{2}$+xy+$y^{2}$)=($x^{3}$+$x^{2}$y+x$y^{2}$)-($x^{2}$y+x$y^{2}$+$y^{3}$)=$x^{3}$-$y^{3}$

This approach will come in handy for at least one problem in chapter two, but I will let you figure out which one!