1. ## Factorisation...

Hi,

This expression can factorised completely.

$\displaystyle (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)$.

Here is how far I've gotten with this.

$\displaystyle (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)$

$\displaystyle =bc^3-b^3c-ac^3+a^3c+ab^3-a^3b$

$\displaystyle =bc(c^2-b^2)-ac(c^2-a^2)+ab(b^2-a^2)$

$\displaystyle =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b-a)$.

Now I need a little push in the right direction. Alternatively, if I've started tackling this wrong a push on where to start it.

Thanks.

2. Originally Posted by bluntpencil
Hi,

This expression can factorised completely.

$\displaystyle (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)$.

Here is how far I've gotten with this.

$\displaystyle (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b)$

$\displaystyle =bc^3-b^3c-ac^3+a^3c+ab^3-a^3b$

$\displaystyle =bc(c^2-b^2)-ac(c^2-a^2)+ab(b^2-a^2)$

$\displaystyle =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b-a)$.

Now I need a little push in the right direction. Alternatively, if I've started tackling this wrong a push on where to start it.

Thanks.
The final step should have been

$\displaystyle =bc(c-b)(c+b)-ac(c-a)(c+a)+ab(b-a)(b\color{red}+\color{black}a)$

3. Argh, damn and blast it. I wrote it out correctly but type it in wrong.

The solution to this is $\displaystyle (a-b)(b-c)(c-a)(a+b+c)$. As you can see I'm alittle bit off of that at this time.

Thanks.