Factoring Cubic Equations: What Methods Can Be Used?

[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}