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 \(a\), \(b\), and \(c\) are the roots of the equation \(x^3 + 2x^2 + 1 = 0\), we can follow these steps:

### Step 1: Identify the coefficients of the cubic equation
The given cubic equation is:
\[ x^3 + 2x^2 + 0x + 1 = 0 \]
Here, we can identify:
- \(a = 1\) (coefficient of \(x^3\))
- \(b = 2\) (coefficient of \(x^2\))
- \(c = 0\) (coefficient of \(x\))
- \(d = 1\) (constant term)

### Step 2: Use Vieta's formulas to find the roots' relationships
From Vieta's formulas, we know:
- The sum of the roots \(a + b + c = -\frac{b}{a} = -\frac{2}{1} = -2\)
- The sum of the product of the roots taken two at a time \(ab + bc + ca = \frac{c}{a} = 0\)
- The product of the roots \(abc = -\frac{d}{a} = -\frac{1}{1} = -1\)

### Step 3: Set up the determinant
We need to calculate the determinant:
\[ D = |[a,b,c],[b,c,a],[c,a,b]| \]

### Step 4: Expand the determinant
Using the determinant formula, we can expand it as follows:
\[ D = a \begin{vmatrix} c & a \\ a & b \end{vmatrix} - b \begin{vmatrix} b & a \\ c & b \end{vmatrix} + c \begin{vmatrix} b & c \\ c & a \end{vmatrix} \]

Calculating each of the 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 back into the determinant:
\[ D = a(cb - a^2) - b(b^2 - ac) + c(ba - c^2) \]
\[ D = acb - a^3 - b^3 + abc + abc - c^3 \]
\[ D = 3abc - (a^3 + b^3 + c^3) \]

### Step 5: Calculate \(a^3 + b^3 + c^3\)
Using the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
We already know:
- \(a + b + c = -2\)
- \(ab + ac + bc = 0\)

Now, we need \(a^2 + b^2 + c^2\):
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) = (-2)^2 - 2(0) = 4 \]
Thus:
\[ a^2 + b^2 + c^2 - ab - ac - bc = 4 - 0 = 4 \]
Now substituting back:
\[ a^3 + b^3 + c^3 - 3(-1) = (-2)(4) \]
\[ a^3 + b^3 + c^3 + 3 = -8 \implies a^3 + b^3 + c^3 = -11 \]

### Step 6: Substitute back into the determinant formula
Now substituting \(abc = -1\) and \(a^3 + b^3 + c^3 = -11\):
\[ D = 3(-1) - (-11) = -3 + 11 = 8 \]

### Final Answer
Thus, the value of the determinant \(|[a,b,c],[b,c,a],[c,a,b]|\) is:
\[ \boxed{8} \]