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

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