if a,b and c are the roots of the equation `x^3+2x^2+1=0`, find `|{:(a,b,x),(b,c,a),(c,a,b):}|`

Options

A. 8
B. `-8`
C. 0
D. 2

Answer

To solve the problem, we need to find the value of the determinant of the matrix formed by the roots \(a\), \(b\), and \(c\) of the polynomial equation \(x^3 + 2x^2 + 1 = 0\). The matrix is given as:

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

### Step 1: Identify the Roots
The roots \(a\), \(b\), and \(c\) can be determined using Vieta's formulas from the polynomial \(x^3 + 2x^2 + 0x + 1 = 0\). According to Vieta's formulas:
- \(a + b + c = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -2\)
- \(ab + bc + ca = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = 0\)
- \(abc = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -1\)

### Step 2: Calculate the Determinant
We can calculate the determinant using the formula for a 3x3 matrix:

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

Using the determinant formula:

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

Calculating each of these 2x2 determinants:

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 these back into the determinant expression:

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

Expanding this gives:

\[ D = acb - a^3 - b^3 + abc + x(ba - c^2) \]

### Step 3: Substitute Roots into the Determinant
Now we substitute \(x = a\), \(x = b\), and \(x = c\) into the determinant expression.

1. For \(x = a\):
\[ D_a = abc - a^3 - b^3 + abc + a(ba - c^2) \]

2. For \(x = b\):
\[ D_b = abc - a^3 - b^3 + abc + b(ba - c^2) \]

3. For \(x = c\):
\[ D_c = abc - a^3 - b^3 + abc + c(ba - c^2) \]

### Step 4: Final Calculation
Using the values from Vieta's:
- \(abc = -1\)
- \(a + b + c = -2\)
- \(ab + ac + bc = 0\)

We can simplify the determinant. The final value of the determinant can be calculated as follows:

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

Using the identity \(a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc)\), we can find \(D\).

After substituting the values, we find:

\[ D = 8 \]

### Conclusion
Thus, the value of the determinant is:

\[ \boxed{8} \]