If a,b and c are the roots of the equation `x^3+2x^2+1=0`, find `|[a,b,c],[b,c,a],[c,a,b]|`.

Options

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

Answer

To find the value of the determinant \( |[a,b,c],[b,c,a],[c,a,b]| \) where \( a, b, c \) are the roots of the polynomial \( x^3 + 2x^2 + 1 = 0 \), we will follow these steps:

### Step 1: Identify the roots' properties
From Vieta's formulas, for the polynomial \( x^3 + 2x^2 + 1 = 0 \):
- The sum of the roots \( a + b + c = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} = -2 \).
- The sum of the product of the roots taken two at a time \( ab + ac + bc = \frac{\text{coefficient of } x}{\text{coefficient of } x^3} = 0 \).
- The product of the roots \( abc = -\frac{\text{constant term}}{\text{coefficient of } x^3} = -1 \).

### Step 2: Set up the determinant
The determinant we need to evaluate is:
\[ D = |[a,b,c],[b,c,a],[c,a,b]| \]

### Step 3: Perform row operations
We can perform the operation \( R_1 \rightarrow R_1 + R_2 + R_3 \):
\[ D = |[a+b+c, a+b+c, a+b+c], [b,c,a], [c,a,b]| \]
This simplifies to:
\[ D = (a+b+c) |[1, 1, 1], [b, c, a], [c, a, b]| \]
Substituting \( a+b+c = -2 \):
\[ D = -2 |[1, 1, 1], [b, c, a], [c, a, b]| \]

### Step 4: Calculate the determinant
Now we need to compute:
\[ |[1, 1, 1], [b, c, a], [c, a, b]| \]
Using the determinant formula for a 3x3 matrix:
\[ = 1 \cdot (ca - ab) - 1 \cdot (cb - ac) + 1 \cdot (ba - bc) \]
This simplifies to:
\[ = ca - ab - cb + ac + ba - bc \]
Rearranging gives:
\[ = ac + ba + ca - (ab + ac + bc) \]
Since \( ab + ac + bc = 0 \):
\[ = ac + ba + ca = 0 \]

### Step 5: Final calculation
Thus, we have:
\[ D = -2 \cdot 0 = 0 \]

### Conclusion
The value of the determinant \( |[a,b,c],[b,c,a],[c,a,b]| \) is:
\[ \boxed{0} \]