If `a, b and c` are distinct and `D = |(a,b,c),(b,c,a),(c,a,b)|.` then the square of the determinant of its cofactors is divisible by

Options

A. `(a^(2)+b^(2)+c^(2))^(2)`
B. `(ab+bc+ca)^(2)`
C. `(a+b+c)^(2)`
D. `(a+b+c)^(4)`

Answer

To solve the problem, we need to find the square of the determinant of the cofactors of the given determinant \( D = |(a,b,c),(b,c,a),(c,a,b)| \) and determine what it is divisible by.

### Step 1: Calculate the Determinant \( D \)

The determinant \( D \) can be calculated using the formula for a 3x3 matrix:

\[ D = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]

Using the determinant formula for a 3x3 matrix, we expand it:

\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating the minors:

1. \( \begin{vmatrix} c & a \\ a & b \end{vmatrix} = cb - a^2 \)

2. \( \begin{vmatrix} b & a \\ c & b \end{vmatrix} = bb - ac = b^2 - ac \)

3. \( \begin{vmatrix} b & c \\ c & a \end{vmatrix} = ba - c^2 \)

Substituting back into the determinant formula:

\[ D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]

Expanding this gives:

\[ D = acb - a^3 - b^3 + abc + abc - c^3 \]

Combining like terms results in:

\[ D = 3abc - (a^3 + b^3 + c^3) \]

### Step 2: Calculate the Square of the Determinant of the Cofactors

The square of the determinant of the cofactors of a matrix \( A \) is given by:

\[ \text{(det A)}^2 = \text{(det A)}^2 \cdot \text{(det I)} = \text{(det A)}^2 \]

For a 3x3 matrix, the determinant of the cofactors is equal to the square of the determinant of the matrix itself. Therefore:

\[ \text{(det of cofactors)}^2 = D^2 \]

### Step 3: Divisibility Condition

We need to determine what \( D^2 \) is divisible by. We already have:

\[ D = 3abc - (a^3 + b^3 + c^3) \]

Using the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

We can express \( D \) as:

\[ D = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

Thus, we can conclude that \( D \) is divisible by \( a + b + c \).

### Final Result

Since \( D^2 \) is divisible by \( (a + b + c)^2 \), we can say that:

\[ D^2 \text{ is divisible by } (a + b + c)^2 \]