Question: Let A be a $3\times 3$ invertible matrix (with real coefficients) and let $B=A^TA^{-1}$. Prove that 

\begin{equation\*}
\det(I + B) = 2(1 + tr(B)).
\end{equation\*}

I know that 

\begin{equation\*}
\det (I+B)=\lambda\_1\lambda\_2\lambda\_3+\lambda\_1\lambda\_2 +\lambda\_1\lambda\_3+\lambda\_2\lambda\_3+\lambda\_1+\lambda\_2+\lambda\_3+1
\end{equation\*}

given $\lambda\_1,\lambda\_2$ and $\lambda\_3$ are three distinct eigenvalues of $B$. However, I don't know where to go on from here and how to utilise the fact that $B=A^TA^{-1}$. Any help or direction would be appreciated.

Answer: 
An alternative approach, which doesn't use the explicit formula in terms of eigenvalues. Instead, find the characteristic polynomial. See that, for any $\lambda$,

$$p(\lambda) = \det (B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \, B - \lambda x + \det B$$

for some number $x$. Here we merely have $\lambda = -1$.

The number $x$ can be computed in a variety of ways. Using exterior algebra is one method: construct the $3 \times 3$ matrix $B\_2$, which acts on $3 \times 1$ column vectors corresponding to elements of $\bigwedge^2 \mathbb R^3$. Then $x = \mathrm{Tr} \, B\_2$.

The relationship between $B\_2$ and $B$ is explicitly

$$B\_2 (a \wedge b) = B(a) \wedge B(b)$$

for any vectors $a, b$.

Now, use a common inversion identity:

$$B^{-1}(a) = \star B\_2^T(\star a)/\det B$$

where $\star$ is the usual Hodge dual. This means we can write $B\_2$ as

$$B\_2(a \wedge b) = \star (B^T)^{-1}(\star [a \wedge b]) \det B$$

You can verify now (e.g. by breaking into a basis) that $\mathrm{Tr} \, B\_2 = \det B \, \mathrm{Tr} \, (B^T)^{-1}$. By the arguments given in other answers, this is merely $\det B \, \mathrm{Tr} \, B$, and as a result, we have

$$p(\lambda) = \det(B - \lambda I) = -\lambda^3 + \lambda^2 \mathrm{Tr} \, B - \lambda [\det B \, \mathrm{Tr} \, B] + \det B$$

For $\lambda =-1$, and since $\det B = 1$, the result follows.
For my opinion, it seems more "comfortable" to use Newton's identities, especially when B is NOT invertible.
e.g.$\lambda\_1\lambda\_2+\lambda\_1\lambda\_3+\lambda\_2\lambda\_3=\dfrac{1}{2} [(\lambda\_1+\lambda\_2+\lambda\_3)^2-(\lambda^2\_1+\lambda^2\_2+\lambda^2\_3)]=\dfrac{1}{2}(tr(B)^2-tr(B^2))$