i have a question and i got stuck on certain step of the solution.
i'll write the question and the solution till the point where i got stuck
and explain why i got stuck

the question:
there is A 3x3 matrices which have 3 real eigenvalues,
so the jordan form of $\displaystyle A^{3} is \left(\begin{array}{ccc}8 & 1 & 0\\0 & 8 & 0\\0 & 0 & 1\end{array}\right)$
find the caracteristic polinomial
and the minimal polinomial of A and write its jordan form.
the solution:
it is given that $\displaystyle A^{3}$is similar to $\displaystyle \left(\begin{array}{ccc}8 & 1 & 0\\0 & 8 & 0\\0 & 0 & 1\end{array}\right)$.
so the caracteristic polinomial of $\displaystyle A^{3}$is
$\displaystyle P_{A}(t)=(t-8)^{2}(t-1)$ so 8 and 1 are the only eigenvalues.
suppose $\displaystyle A\in M_{3x3}^{C}$ and $\displaystyle a,b,c\in C$ so $\displaystyle P_{A}(t)=(t-a)(t-b)(t-c)$
if $\displaystyle \lambda\in C$ and its eigenvalue of A then $\displaystyle \lambda^{3}$ is eigen value of $\displaystyle A^{3}$
so $\displaystyle \{a^{3},b^{3},c^{3}\}\subseteq\{1,8\}$
so $\displaystyle \{a,b,c\}\subseteq\{1,2\}$
so the possible caracteristic polinomials are:
$\displaystyle (t-2)^{3},(t-1)^{3},(t-2)^{2}(t-1),(t-2)(t-1)^{2}$
suppose that $\displaystyle P_{A}(t)=(t-2)^{3}$ or $\displaystyle P_{A}(t)=(t-1)^{3}$
then the caracteristic polinomial will be $\displaystyle P_{A}(t)=(t-2^{3})^{3}$ or
$\displaystyle P_{A}(t)=(t-1^{3})^{3}$and thats cannot happen .(stuck on understanding this point)

my probem with this point:
if $\displaystyle P_{A}(t)=(t-2)^{3}$ i know that 8 is eigenvalue of $\displaystyle A^{3}$ but why the $\displaystyle P_{A^{3}}(t)=(t-8)^{3}$
why we have the same structure??
another problem is why these possible caracteristic polinomials cannot work?