<s>[INST] 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?
 [/INST] Observe you have
\begin{align}
r^6-3r^4+3r^2-1 = (r^2-1)^3
\end{align}
by the binomial theorem. Then it follows $(r^2-1) = (r+1)(r-1)$. 
</s>