How to factor $r^6 -3r^4 +3r^2 - 1 = 0$ I tried two ways to factor $r^6 -3r^4 +3r^2 - 1 = 0$
When I factor $r^4$ out of $r^6 -3r^4$:
$r^4(r^2-3)+(3r^2-1) = 0$
When I factor $r^2$ out of $r^6 + 3r^2$:
$r^2(r^4+3)-(3r^4 + 1) = 0$
For both methods, I'm stuck at an equation that isn't factorable. But Wolfram Alpha says $r^6 -3r^4 +3r^2 - 1 = (r-1)^3(r+1)^3$. How do you get this?

Option 1
Notice that if you make the substitution $x = r^2$ to make it $x^3-3x^2+3x-1$, the coefficients are (1, -3, 3, -1), which is exactly the values on the 4th row of Pascal's triangle, with alternating signs. So it must be something like $(x \pm 1)^3$. The last coefficient is $-1$, so there must be an odd number of $(x-1)$ terms, and it doesn't take long to realise it must be $(x-1)^3=(r^2-1)^3=(r-1)^3(r+1)^3$.
Option 2
Using the rational roots theorem, we note that both $r=1$ and $r=-1$ are roots. Divide the whole thing by $(r^2-1)$ and we get $r^4-2r^2+1$. Applying rational roots again and we see that $r=1$ and $r=-1$ are still roots. Divide again by $(r^2-1)$ and we're left with $r^2-1$, so we know that the whole thing is $(r^2-1)^3=(r-1)^3(r+1)^3$.
