<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] If you don't know the binomial theorem, you might rearrange the terms:
\begin{align*}
  r^6-3r^4+3r^2-1 &= r^6-1-3(r^4-r^2) \\
  &= (r^2-1)(r^4+r^2+1)-3r^2(r^2-1) \\
  &= (r^2-1)(r^4+r^2+1-3r^2) \\
  &= (r^2-1)(r^4-2r^2+1) \\
  &= (r^2-1)(r^2-1)^2 \\
  &= (r^2-1)^3 \\
  &= (r+1)^3(r-1)^3
\end{align*}

Useful identities:
  \begin{align*}
  a^2 \pm 2ab+b^2&= (a \pm b)^2 \\
  a^2-b^2 &= (a+b)(a-b) \\
 a^3 \pm b^3 &= (a \pm b)(a^2 \mp ab + b^2)
\end{align*}

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