factoRRRR this...and I don't know where to start
$\displaystyle y^6-1$
Do you mean this??
$\displaystyle = (y^2)^3 - 1^3$
Where $\displaystyle a=y^2$ and $\displaystyle b=1$
$\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$
Then we have...
$\displaystyle (y^2)^3-1^3=(y^2-1) [(y^2)^2+y^2+1)]$
$\displaystyle (y^2)^3-1^3=(y^2-1) (y^4+y^2+1)$
...But I don't think this is the answer...mmm...I really don't
How can you possibly think I mean that when I clearly wrote this:
Now all you have to do is factorise each of $\displaystyle y^3 - 1 = y^3 - 1^3$ and $\displaystyle y^3 + 1 = y^3 + 1^3$.$\displaystyle = (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....$
And you should know how to factorise the sum and difference of two perfect cubes.