How do I factor x^3-1?

[candynrg]

How do I factor this equation
x^3-1

 

[Pyrrhus]

Difference of cubes you mean

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

 

[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.

 

[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

 

[whozum]

It does work

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

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

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

 

[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

 

[z-component]

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

 

[maverickmathematics]

z-component I was making reference to

No cubes there,just squares.

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

 

[z-component]

Ah, ok. I understand.

 

[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...

 

[HallsofIvy]

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- y^{n= (x-y)(xn-1+ xn-2y+ . . .+ x2yn-3+ xyn-2+ yn-1).}

 

[roger]

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- y^{n= (x-y)(xn-1+ xn-2y+ . . .+ x2yn-3+ xyn-2+ yn-1).}

^{how is that last general form, derived ?}

 

[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.

 

[robert Ihnot]

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

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

 

[Curious3141]

how is that last general form, derived ?

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 $$\frac{y^n - x^n}{y - x}$$