If a,b,c are the roots of `x^3+px^2+q=0` , then the value of `|(a,b,c),(b,c,a),(c,a,b)|` is equal to

Options

A. `-p^3`
B. `p^3-3q`
C. `p^3`
D. `p^2-3q`

Answer

To solve the problem, we need to find the value of the determinant:

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

where \( a, b, c \) are the roots of the polynomial \( x^3 + px^2 + q = 0 \).

### Step 1: Write the determinant

We start with the determinant:

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

### Step 2: Use properties of determinants

We can use the property of determinants that allows us to add rows. Let's add all three rows together to the first row:

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

### Step 3: Factor out the common term

Now, we can factor out \( a + b + c \) from the first row:

\[ D = (a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]

### Step 4: Calculate the smaller determinant

Next, we need to calculate the smaller determinant:

\[ D' = \begin{vmatrix} 1 & 1 & 1 \\ b & c & a \\ c & a & b \end{vmatrix} \]

Using the determinant formula, we can expand this:

\[ D' = 1 \cdot \begin{vmatrix} c & a \\ a & b \end{vmatrix} - 1 \cdot \begin{vmatrix} b & a \\ c & b \end{vmatrix} + 1 \cdot \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of these 2x2 determinants, we have:

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

Thus, we have:

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

### Step 5: Simplify \( D' \)

Combining these terms:

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

### Step 6: Substitute back into \( D \)

Now, substituting back into \( D \):

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

### Step 7: Use the relationships from the polynomial

From the polynomial \( x^3 + px^2 + q = 0 \), we know:

- \( a + b + c = -p \)
- \( ab + ac + bc = 0 \) (since the coefficient of \( x \) is 0)
- \( abc = -q \)

### Step 8: Final simplification

Using the relationships, we can further simplify \( D \):

Since \( ab + ac + bc = 0 \), we can substitute into our expression for \( D' \) to find that the determinant simplifies to 0.

Thus, the final value of the determinant is:

\[ D = 0 \]

### Conclusion

The value of the determinant \( |(a,b,c),(b,c,a),(c,a,b)| \) is equal to \( 0 \).