Determinant of a $4 \times 4$ matrix $A$ and $(\det(A))^5$ Calculate $\det(A)$ and $\det(A)^5$:
$$A= \begin{bmatrix}a&a&a&a\\a&b&b&b\\a&b&c&c\\a&b&c&d\end{bmatrix}$$
I found $\det A$ with Laplace expansion: $$a(-abc+b^2c+a c^2-b c^2+a b d-b^2 d-a c d+b c d) .$$
But how can I determine $\det A^5$ easily/fast? I know that there is the possibility to use $\det(A^5)=\det(A)^5)$, but this is too long for given time to resolve the problem.

If calculations are done mentally but not jotted down, I think the quickest way to calculate $\det A$ is to note that
\begin{align}
A&=a(e_1+e_2+e_3+e_4)(e_1+e_2+e_3+e_4)^T\\
&+(b-a)(e_2+e_3+e_4)(e_2+e_3+e_4)^T\\
&+(c-b)(e_3+e_4)(e_3+e_4)^T\\
&+(d-c)e_4e_4^T.
\end{align}
Therefore $A$ has the $LDL^T$ decomposition
$$
A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1}
\pmatrix{a\\ &b-a\\ &&c-b\\ &&&d-c}
\pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1}
$$
and $\det A=a(b-a)(c-b)(d-c)$. And as noted by the others, $\det A^5$ is just $(\det A)^5$.
The above decomposition has the additional merit that it gives also the signature of $A$. In particular, $A$ and $A^5$ are positive semidefinite if and only if $d\ge c\ge b\ge a\ge0$.
