Does not algebraic multiplicity= geometric multiplicity $\Rightarrow$ the matrix is diagonalizable? 
let $A=\left(\begin{array}{ccc} -5 & -1 & 6 \\ -2 & -5 & 8 \\ -1 & -1 & 1 \end{array}\right)$ and $B=\left(\begin{array}{ccc} -9 & 3 & -3 \\ -14 & 4 & -7 \\ -2 & 1 & -4 \end{array}\right)$ matrices above $\mathbb{C}$ are they similar?

So I started with finding the eigenvalues and geometric multiplicity.
For $A$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^3$
For $B$ I got $f_{\lambda}(x)=(x+3)^3$ and $m_{\lambda}(x)=(x+3)^2$
So the jordan normal form of $A$ is $A=J_{1}(-3),J_{1}(-3),J_{1}(-3)$ or $A=\left(\begin{array}{ccc} -3 & 0 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
And  the jordan normal form of $B$ is $B=J_{2}(-3),J_{1}(-3)$  or $B=\left(\begin{array}{ccc} -3 & 1 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -3 \end{array}\right)$
So $A$ is not similar to $B$ is it correct?

Indeed they are not similar. But $A$ is in fact not diagolizable.
It has a size 3 Jordan block:
$(A+3E)^2 \neq 0$ (whereas $(B+3E)^2=0$). This is also what you got using the $m_\lambda$ formulation.
