<s>[INST] 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.
 [/INST] @Michael Freimann 
What follows is not an answer. It is just an aside comment, that might interest some readers.
The kind of matrices considered in this question can be described by the following formula for their coefficients 
$$A_{ij}=a_{min(i,j)}$$ for a given sequence $a_1,a_2,...a_n$, this sequence being here {a,b,c,d}.
Their inverses have a tridiagonal form with simple coefficients. For example, here:
$A^{-1}=\begin{pmatrix} \frac{1}{a} + \frac{1}{b-a}&\frac{1}{a - b}&0&0\\
\frac{1}{a - b}&\frac{1}{b-a} + \frac{1}{c-b}&\frac{1}{b - c}&0\\
0&\frac{1}{b - c}&\frac{1}{c-b} + \frac{1}{d-c }&\frac{1}{c - d}\\
0&0&\frac{1}{c - d}&\frac{1}{d-c}\end{pmatrix}$
In particular, when $a=b=c=d=1$, 
$$\text{The inverse of} \ \ \begin{pmatrix}
1&1&1&1\\
1&2&2&2\\
1&2&3&3\\
1&2&3&4
\end{pmatrix} \ \  \text{is} \ \ 
\frac{1}{2}\begin{pmatrix} 2&-1&0&0\\
-1&2&-1&0\\
0&-1&2&-1\\
0&0&-1&1\end{pmatrix}
$$
One can recognized in the last matrix an approximation of the opposite of the second derivative, a very important matrix in numerical analysis: see this (slides 10, 29, 35, 37, 42...).
This is in clear connection with the decomposition given by @user1551, which, for $a=c=d=1$, gives: 
$$A=\pmatrix{1&0&0&0\\ 1&1&0&0\\ 1&1&1&0\\ 1&1&1&1}
\pmatrix{1&1&1&1\\ 0&1&1&1\\ 0&0&1&1\\ 0&0&0&1}$$
which is equivalent to a double discrete integration operator.
Of course, this property extends naturally to any dimension. Moreover, it has a nice correspondence with the "continuous world": see this.
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