<s>[INST] 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.
 [/INST] HINT : Since we know $f(a)=a^3-3a+1=0\iff a^3=3a-1,$ use this to prove $$f(a^2-2)=0.$$
$$f(a^2-2)=(a^2-2)^3-3(a^2-2)+1=a^6-6a^4+9a^2-1$$$$=(a^3-3a+1)^2-2(a^3-3a+1)=0^2-2\cdot 0=0.$$
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