Answer
See below
Work Step by Step
Suppose A is a $6 \times 6$ matrix with eigenvalue $\lambda$ (of multiplicity 6).
Given that $(A-\lambda I)^3=0$ and $(A-\lambda I)^2 \ne 0 $
Then we know that there will be 3 possibilities here:
$I_1=\begin{bmatrix}
\lambda & 1 & 0 & 0 & 0 & 0\\
0 & \lambda & 1 & 0 & 0 & 0\\
0 & 0 & \lambda & 0 & 0 & 0\\
0 & 0 & 0 & \lambda & 1 & 0 \\
0 & 0 & 0 & 0 & \lambda & 1 \\
0 & 0 & 0 & 0 & 0 & \lambda
\end{bmatrix}$
$I_2=\begin{bmatrix}
\lambda & 1 & 0 & 0 & 0 & 0\\
0 & \lambda & 1 & 0 & 0 & 0\\
0 & 0 & \lambda & 0 & 0 & 0\\
0 & 0 & 0 & \lambda & 1 & 0 \\
0 & 0 & 0 & 0 & \lambda & 0 \\
0 & 0 & 0 & 0 & 0 & \lambda
\end{bmatrix}$
$I_3=\begin{bmatrix}
\lambda & 1 & 0 & 0 & 0 & 0\\
0 & \lambda & 1 & 0 & 0 & 0\\
0 & 0 & \lambda & 0 & 0 & 0\\
0 & 0 & 0 & \lambda & 0 & 0 \\
0 & 0 & 0 & 0 & \lambda & 0 \\
0 & 0 & 0 & 0 & 0 & \lambda
\end{bmatrix}$
Hence, there are three possible Jordan canonical forms of $A$
