`|(1,a,a^2,a^3+bcd),(1,b,b^2,b^3+cda),(1,c,c^2,c^3+abd),(1,d,d^2,d^3+abc)|` =

Options

A. `a^3b^3c^3`
B. `b^3c^3d^3`
C. `c^3d^3a^3`
D. 0

Answer

To solve the determinant

\[ D = \begin{vmatrix} 1 & a & a^2 & a^3 + bcd \\ 1 & b & b^2 & b^3 + cda \\ 1 & c & c^2 & c^3 + abd \\ 1 & d & d^2 & d^3 + abc \end{vmatrix} \]

we can use properties of determinants to simplify the calculation.

### Step 1: Split the determinant
We can split the last column into two parts:

\[ D = \begin{vmatrix} 1 & a & a^2 & a^3 \\ 1 & b & b^2 & b^3 \\ 1 & c & c^2 & c^3 \\ 1 & d & d^2 & d^3 \end{vmatrix} + \begin{vmatrix} 1 & a & a^2 & bcd \\ 1 & b & b^2 & cda \\ 1 & c & c^2 & abd \\ 1 & d & d^2 & abc \end{vmatrix} \]

Let's denote the first determinant as \(D_1\) and the second as \(D_2\).

### Step 2: Evaluate \(D_1\)
The determinant \(D_1\) is a Vandermonde determinant, which can be calculated as:

\[ D_1 = (b-a)(c-a)(d-a)(c-b)(d-b)(d-c) \]

### Step 3: Evaluate \(D_2\)
Now, we will evaluate \(D_2\). Notice that we can factor out \(bcd\), \(cda\), \(abd\), and \(abc\) from the last column.

We can express \(D_2\) as:

\[ D_2 = \begin{vmatrix} 1 & a & a^2 & 0 \\ 1 & b & b^2 & 0 \\ 1 & c & c^2 & 0 \\ 1 & d & d^2 & 0 \end{vmatrix} + \begin{vmatrix} 1 & a & a^2 & bcd \\ 1 & b & b^2 & cda \\ 1 & c & c^2 & abd \\ 1 & d & d^2 & abc \end{vmatrix} \]

The first part is clearly zero since the last column is all zeros. Therefore, \(D_2\) simplifies to:

\[ D_2 = \begin{vmatrix} 1 & a & a^2 & bcd \\ 1 & b & b^2 & cda \\ 1 & c & c^2 & abd \\ 1 & d & d^2 & abc \end{vmatrix} \]

### Step 4: Combine \(D_1\) and \(D_2\)
Now, we can combine \(D_1\) and \(D_2\):

\[ D = D_1 + D_2 \]

### Step 5: Final evaluation
Since \(D_1\) is non-zero (as long as \(a\), \(b\), \(c\), and \(d\) are distinct), and \(D_2\) can be shown to be zero due to the properties of determinants (as the last column can be expressed as a linear combination of the other columns), we conclude that:

\[ D = D_1 + 0 = D_1 \]

Thus, the final value of the determinant \(D\) is:

\[ D = 0 \]