I can prove this statement wrong by considering $A\neq\mathbf O$ and $A$ not diagonalizable while $A^3=\mathbf O$.
But what's wrong with my intuitive proof:
$A^3=PDP^{-1}$, and we just consider $A=PD^{1/3}P^{-1}$, it seems that everything is satisfied. And furthermore, we can also show uniqueness by considering in certain eigenspace $V_\lambda$, $A^3=\lambda I$ so the eigenvalue of $A$ must be $\sqrt[3]\lambda$.
Could anyone help me to figure out what's wrong here? And why does the converse hold?
Let us consider the problem in the field $\mathbb{C}$ of complex numbers.
If $0$ is an eigenvalue of $A^3$, the result is false in general as the following example shows \begin{equation} A = \begin{pmatrix}0&1&0\cr 0&0&1\cr 0&0&0\cr\end{pmatrix} \qquad A^3=0 \end{equation}
If $0$ is not an eigenvalue of $A^3$, the result is true by the following argument: assuming $A^3$ is diagonalizable, let $P(X) = \prod_i (X-\mu_i)$ be the minimal polynomial of $A^3$, with $\mu_i\not = \mu_j$ for $i\not = j$, and let $Q(x)=\prod (X-\lambda_i)^{m_i}$ be the minimal polynomial of $A$. Obviously, \begin{equation} R(X) = P(X^3) = \prod_i (X^3-\mu_i) \end{equation} is a multiple of $Q(X)$ in $\mathbb{C}[X]$ because $R(A)=0$. It follows that if $\lambda$ is a root of $Q$, then $R(\lambda) = 0$ and one has $\lambda^3 = \mu_k$ for some index $k$. In particular one has $\lambda\not = 0$. If $\lambda$ is a multiple root of $Q$, then it must also be a multiple root of $R$. But one has \begin{equation} R'(\lambda) = \sum_i 3 \lambda^2 \prod_{j\not = i} (\lambda^3-\mu_j) = 3 \lambda^2 \prod_{j\not = k} (\mu_k-\mu_j) \not = 0 \end{equation} Thus $\lambda$ cannot be a multiple root of $R$, hence $Q$ has only simple roots and $A$ is diagonalizable.