# Thread: Need to facto this.

1. ## Need to facto this.

factoRRRR this...and I don't know where to start

$y^6-1$

2. Originally Posted by Alienis Back
factoRRRR this...and I don't know where to start

$y^6-1$
$= (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....$

And you should know how to factorise the sum and difference of two perfect cubes.

3. Do you mean this??

$= (y^2)^3 - 1^3$

Where $a=y^2$ and $b=1$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then we have...

$(y^2)^3-1^3=(y^2-1) [(y^2)^2+y^2+1)]$

$(y^2)^3-1^3=(y^2-1) (y^4+y^2+1)$

...But I don't think this is the answer...mmm...I really don't

4. Originally Posted by Alienis Back
Do you mean this??

$= (y^2)^3 - 1^3$

Where $a=y^2$ and $b=1$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

[snip]
How can you possibly think I mean that when I clearly wrote this:

$= (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....$

And you should know how to factorise the sum and difference of two perfect cubes.
Now all you have to do is factorise each of $y^3 - 1 = y^3 - 1^3$ and $y^3 + 1 = y^3 + 1^3$.