I was working on the following problem when I stumbled upon an oddity.
If $X=P^{-1}AP$ and $A^3=I$, prove that $X^3=I$
My first approach was to cube both side which led to the following:
$$X^3=(P^{-1}AP)^3$$ $$X^3=(P^{-1})^3A^3P^3$$ Since $A^3=I$ $$X^3=(P^{-1})^3I^3P^3$$ $$X^3=(P^{-1})^3P^3$$ $$X^3=(P^{-1}P)^3$$ $$X=I^3=I$$ However this method seemed to contradict what I had previously learned. Instead I tried this approach: $$X=P^{-1}AP$$ Now premultiply each side by X and we get: $$X^2=P^{-1}APP^{-1}AP=P^{-1}AIAP=P^{-1}A^2P$$
We do this once more: $$X^3=P^{-1}A^2PP^{-1}AP=P^{-1}A^2IAP=P^{-1}A^3P$$
Since $A^3=I$ $$X^3=P^{-1}IP=P^{-1}P=I$$
Does anyone know which approach is correct?