Finding $\det(-A^3B^TA^{-1})$ given $\det(A)$ and $\det(B)$ This is my first question, please forgive me if I mistake something.
The question I have is; 

Let A and B are square matrices of order 5. Find $\det(-A^3B^TA^{-1})$ while $ \det(A)=5 $ and $ \det(B)=1/8 $

What I did was put everything into place knowing $ \det(B)^T=\det(B)$ and what I got is something like this;

$ \det((-125)(1/8)(1/5))\Rightarrow \det(-125/40) \Rightarrow 5(-125/40) \Rightarrow -125/8 $

but I am concerned that my knowledge here is not certain thus there is probably something I am missing. I would very much appreciate if you can just clear my mind on this by pointing what's wrong if there is anything. Again thank you.
EDIT:
With the information from the comments I came to this conclusion;
$ (-125)(1/8)(1/5) $ = $ -25/8 $
But what does the information of being order of five give me in this question if I don't need to factor it with five at the end? Is it just for the  $ \det(cA)=c^n\det(A)  $?
 A: If $A$ is an $n\times n$ matrix, then
$$
\det(-A)=(-1)^n\det A
$$
because of the more general result you cite: $\det(cA)=c^n\det(A)$, with $c=-1$.
This follows from the fact that, denoting by $I$ the identity matrix, we have $cA=(cI)A$ and so Binet's theorem says that $\det(cA)=\det((cI)A)=\det(cI)\det(A)$. Now $cI$ is a diagonal matrix, so its determinant is the product of the entries along the diagonal, which are all equal to $c$ and there are $n$ of them. Thus $\det(cI)=c^n$.
In your case $n=5$, so you can say that
$$
\det(-A^3B^TA^{-1})=-\det(A^3B^TA^{-1})
$$
By Binet's theorem and taking into account that $\det(B^T)=\det(B)$ and $\det(A^{-1})=\det(A)^{-1}$,
\begin{align}
\det(-A^3B^TA^{-1})
&=-\det(A^3B^TA^{-1})\\[4px]
&=-\det(A^3)\det(B^T)\det(A^{-1})\\[4px]
&=-\det(A)^3\det(B)\det(A)^{-1}\\[4px]
&=-5^3\cdot\frac{1}{8}\cdot\frac{1}{5}=-\frac{25}{8}
\end{align}
There is no “multiplication by $5$” to do.
A: By the properties of determinants, the determinant you're calculating is
$$\det(-A^3B^TA^{-1}) = (-1) \times det(A^3) \times det(B^T) \times det A^{-1} = (-1) \times det(A^2) \times det(B) $$ $$ = (-1) \times (det(A))^2 \times det(B) = \frac{-25}{8}$$
Notice that $$det(A^{-1}) = \frac{1}{det(A)}$$ and $det(A^3) = det(AAA) = det(A) \times det(A) \times det(A) $
