Difference of squares

1. May 17, 2005

candynrg

How do I factor this equation
x^3-1

2. May 17, 2005

Pyrrhus

Difference of cubes you mean

$$(x - 1)(x^2 + x + 1)$$

3. May 17, 2005

dextercioby

No cubes there,just squares.

$$\left(x^{\frac{3}{2}}\right)^{2}-1^{2}$$

Daniel.

P.S.Regarding a proof for the formula Cyclovenom posted,try polynomial division.

4. May 17, 2005

maverickmathematics

Isn't $$x \times x^2 = x^3$$ a cube? Or is the power of 3 now a square?!

Cyclovenom's factorisation $$(x - 1)(x^2 + x + 1)$$ works fine and the problem is typically the difference of two cubes. Your factorisation works fine too.

-M

Last edited: May 17, 2005
5. May 17, 2005

whozum

It does work

$$a^2 - b^2 = (a+b)(a-b)$$

$$a = x^{\frac{3}{2}}, b = 1$$

$$(x^\frac{3}{2} + 1)(x^\frac{3}{2}-1)$$

6. May 17, 2005

maverickmathematics

Hey i was reading something on paper and was thinking in about 5 dimensions and it isn't always easy for me to come back down to one - so sorry.

I corrected the erratum a few minutes later when I begn thinking about complex numbers and the post now stands happy and correct.

The mistake i made was not treating the statement as
$$- 1^2$$ which equals $$-1$$

but as
$$(- 1)^2$$ which of course equals $$1$$

-M

Last edited: May 17, 2005
7. May 17, 2005

z-component

Well yes $$x^3$$ is certainly a cube, but Daniel is saying that $$\left(x^{\frac{3}{2}}\right)^{2}=x^3$$.

8. May 17, 2005

maverickmathematics

z-component I was making reference to

when he was talking about the difference of two cubes.

i of course was not trying to deny the fact that $$\left(x^{\frac{3}{2}}\right)^{2}=x^3$$

-M

9. May 17, 2005

z-component

Ah, ok. I understand.

10. May 17, 2005

Zurtex

Read the title of the thread "difference of squares", which would imply it's asking how to factor my difference of squares as apposed to any other method. It's probabily a mistake by the author but still...

11. May 17, 2005

HallsofIvy

Staff Emeritus
However, to get back to the question posed before dextercioby threw his monkey wrench into the machinery: x3- 1 is a sum of cubes and can be factored as x3- 1= (x-1)(x2+ x+ 1). In general xn- yn= (x-y)(xn-1+ xn-2y+ . . .+ x2yn-3+ xyn-2+ yn-1).

12. May 18, 2005

roger

how is that last general form, derived ?

13. May 18, 2005

dextercioby

Polynomial division.You know that the polynomial $P(x)=x^{n}-y^{n} \ ,\mbox{n=odd}$ has at least one real root and it's easy to see that the root is "y".Therefore you know that the remainder of the division of the polynomial $P(x)$ through the monom $x-y$ is zero and all u have to do is compute the ratio.

Daniel.

14. May 18, 2005

robert Ihnot

May I add, lettng $$\zeta =\frac{-1+\sqrt-3}{2}$$, we can continue the factorization arriving at:

$$X^3-Y^3 = (X-Y)(X-\zeta{Y})(X-\zeta^2Y)$$

Last edited: May 18, 2005
15. May 18, 2005

Curious3141

Polynomial division is one way. The other is using the formula for sum of a geometric progression.

Note that the series that HallsofIvy posted is a geometric sum with first term $x^{(n-1)}$ and common ratio $\frac{y}{x}$ giving the sum (to n terms) of :

$$\frac{x^{n-1}({(\frac{y}{x})}^n - 1)}{\frac{y}{x} - 1}$$ which reduces to $$\frac{y^n - x^n}{y - x}$$