# Factoring cubic equation

1. Jan 2, 2008

### bizkut

Factorising cubic equation

Anyone here know how to factor this equation?

1. The problem statement, all variables and given/known data

$$a^{3}c-a^{3}b+b^{3}a-b^{3}c+c^{3}b-c^{3}a$$

3. The attempt at a solution

I tried factoring by grouping but ended up getting nowhere.

If anyone can factor this equation, please tell me step by step how you got it.

Last edited: Jan 2, 2008
2. Jan 3, 2008

### HallsofIvy

Staff Emeritus
Cubics are bad enough but this has 3 variables. This is not a complete factorization but I think the most simplified form is
$$(a^3c- ac^3)+ (b^3a- a^3b)+ (c^3b- b^3c)= ac(a^2- c^2)+ ab(b^2- a^2)+ bc(c^2- b^2)$$
$$= ac(a- c)(a+ c)+ ab(b- a)(b+ a)+ bd(c- b)(c+ b)$$

3. Jan 4, 2008

### bizkut

thanks, that will be ok.

but if you could figure how to get this form, please let me know.
$$(a+b+c)(b-c)(c-a)(a-b)$$

4. Jan 5, 2008

### Gib Z

Notice the symmetry in the variables. Then, without loss of generality, treat it as a polynomial in a, and collect coefficients.

5. Jan 5, 2008

### Feldoh

Yeah, I'd have to agree with Gib Z