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?
 A: 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 $(r^2-1) = (r+1)(r-1)$. 
A: Let $x = r^{2}$
Then $x^{3} -3x^2+3x-1 = 0$
Notice that $x=1$ is a solution
Therefore we have $(x-1)(x^{2} +ax+1) = 0$ where $a$ is a constant we need to find
By comparing coefficients of $x$, we can see that $-a+1 = 3 \Rightarrow a = -2$
Then we have $(x-1)(x^2-2x+1)=0 \Rightarrow (x-1)(x-1)^{2}=0 \Rightarrow (x-1)^{3} = 0$
Then $(r^2-1)^3=0$
Alternatively, you could notice this is the binomial expansion (comes with experience) and thus factorise by inspection.
A: 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*}

A: 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 $(r^2-1)^3=(r-1)^3(r+1)^3$.
