On the roots of a polynomial Let $f(x)= x^3 - 3x + 1$. Show that if a complex number $a$ is root of $f(x)$, then $a^2-2$ is too.
 A: HINT : Since we know $f(a)=a^3-3a+1=0\iff a^3=3a-1,$ use this to prove $$f(a^2-2)=0.$$
$$f(a^2-2)=(a^2-2)^3-3(a^2-2)+1=a^6-6a^4+9a^2-1$$$$=(a^3-3a+1)^2-2(a^3-3a+1)=0^2-2\cdot 0=0.$$
A: Compute $f(a^2-2)$ using simple ring properties and observe it is equal to $f(a)g(a)$ for some polynomial $g(a)$.
A: Hint $\,\ f(x) = - x^3 f(1\!-\!x^{-1}),\,$ and $\,  1\!-\!x^{-1} = x^2\!-2\,$ if $\,f(x) = x^3-2x+1 = 0$
A: It reminds me of $\displaystyle \cos3\phi=4\cos^3\phi-3\cos\phi$
$\displaystyle x^3-3x+1=0\implies 4\left(\frac x2\right)^3-3\left(\frac x2\right)=-\frac12\  \ \ \ (1)$
If we set $\displaystyle\frac x2=\cos\phi,\cos3\phi=-\frac12=\cos120^\circ$
$\displaystyle\implies3\phi=360^\circ n\pm120^\circ\iff\phi=120^\circ n\pm40^\circ,$ where $n$ is any integer
So, the roots of $(1)$ are $\displaystyle2\cos40^\circ,2\cos80^\circ,2\cos160^\circ$
Now check using $\displaystyle\cos2A=2\cos^2A-1\iff 2\cos^2A=1+\cos2A$
For example, if $\displaystyle a=2\cos160^\circ,$
$\displaystyle a^2-2=(2\cos160^\circ)^2-2=2(1+\cos(2\cdot160^\circ)+1)-2$
$\displaystyle=2\cos320^\circ=2\cos(360-40)^\circ=2\cos40^\circ$ and so on
