<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] 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.
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