$A\in M\_3(\mathbb R)$ orthogonal matrix with $\det (A)=1$.

I need to prove that $(\mathrm{tr} A)^2-\mathrm{tr}(A^2) = 2 \mathrm{tr} (A)$ ; $\mathrm{tr}$=trace.

I know that if $A$ is orthogonal than $A^tA=I$ and that $A$ is diagonalizable and similar to $D=\begin{pmatrix}
\lambda\_1 & & \\ 
 & \lambda\_2 & \\ 
 & & \lambda\_3
\end{pmatrix}$. We know as well that $\mathrm{tr} A=\mathrm{tr} D= \lambda\_1+ \lambda\_2+\lambda\_3$ that $\mathrm{tr} A^2=\mathrm{tr} D^2= \lambda\_1^2+ \lambda\_2^2+\lambda\_3^2$ and that 
$\lambda\_1 \lambda\_2 \lambda\_3=1$. It's not enough for solving the question.

What more should I know or use in order to solve it?

Thanks