EigenValue and Basis Problem 
The matrix
  $A=\begin{bmatrix}0 & 3 & 3 \\ -3 & -6 & -3 \\ 3 & 3 & 0 \end{bmatrix}$ has two real eigenvalues, one of multiplicity $1$ and one of multiplicity $2$. Find the eigenvalues and a basis of each eigenspace.

I have found that $\det(A−λI)= -λ^3 -6λ^2 -27$. But I'm having a hard time factoring this cubic equation. Any hints?
 A: Notice that $$R_1+R_3=-R_2$$
where $R_i$ denotes the $i$-th row of the matrix. Hence $0$ must be an eigenvalue and you must have made some computational mistake.
Also, $R_1$ and $R_3$ are not scalar multiple of each other, hence the rank of matrix is at least $2$. 
Hence multiplicity of eigenvalue $0$ must be $1$ and we have another eigenvalue, $\alpha$ with multiplicty $2$.
$$tr(A)=-6$$
$$0+2\alpha = -6$$
$$\alpha = -3$$
A: Sometimes the characterist equation just gets in the way.
$A = \begin{bmatrix} 0 &3 &3\\-3&-6&-3\\3&3&0\end{bmatrix}$
The 1 column + 3rd colum - second column produces a zero vector.
$A$ is singular.
$\det (A) = 0.$
$0$ is an eigenvalue.
$\begin{bmatrix} 1\\-1\\1 \end{bmatrix}$ is an eigenvector
$(A-\lambda I) = \begin{bmatrix} -\lambda &3 &3\\-3&-6-\lambda&-3\\3&3&-\lambda\end{bmatrix}$
Just looking at the first and 3rd rows.
$\lambda = -3$ is going to make those two rows equal and hence 
$\det(A + 3I) = 0$
$(A+3I) = \begin{bmatrix} 3 &3 &3\\-3&-3&-3\\3&3&3\end{bmatrix}$
$\begin{bmatrix} 1\\-1\\0 \end{bmatrix},\begin{bmatrix} 0\\-1\\1 \end{bmatrix}$  are eigenvectors
