If roots of `x ^(3) +2x ^(2) +1=0` are `alpha, beta and gamma,` then the vlaue of `(alpha beta)^(3) + (beta gamma )^(3) + (alpha gamma )^(3) ,` is :

Options

A. `-11`
B. 3
C. 0
D. `-2`

Answer

To solve the problem, we need to find the value of \((\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3\) given the cubic equation \(x^3 + 2x^2 + 1 = 0\) with roots \(\alpha, \beta, \gamma\).

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

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \(ax^3 + bx^2 + cx + d = 0\):
- The sum of the roots \(\alpha + \beta + \gamma = -\frac{b}{a} = -\frac{2}{1} = -2\)
- The sum of the products of the roots taken two at a time \(\alpha \beta + \beta \gamma + \alpha \gamma = \frac{c}{a} = \frac{0}{1} = 0\)
- The product of the roots \(\alpha \beta \gamma = -\frac{d}{a} = -\frac{1}{1} = -1\)

### Step 3: Calculate \((\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3\)
We can use the identity for the sum of cubes:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
Let:
- \(a = \alpha \beta\)
- \(b = \beta \gamma\)
- \(c = \alpha \gamma\)

From Vieta's, we know:
\[ \alpha \beta + \beta \gamma + \alpha \gamma = 0 \]
Thus, \(a + b + c = 0\).

### Step 4: Substitute into the identity
Substituting into the identity:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 - 3(\alpha \beta)(\beta \gamma)(\alpha \gamma) = 0 \]
This simplifies to:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 = 3(\alpha \beta)(\beta \gamma)(\alpha \gamma) \]

### Step 5: Calculate \((\alpha \beta)(\beta \gamma)(\alpha \gamma)\)
We know:
\[ \alpha \beta \gamma = -1 \]
Thus:
\[ (\alpha \beta)(\beta \gamma)(\alpha \gamma) = \alpha^2 \beta^2 \gamma^2 = (\alpha \beta \gamma)^2 = (-1)^2 = 1 \]

### Step 6: Final calculation
Substituting back, we get:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 = 3 \times 1 = 3 \]

### Conclusion
The value of \((\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3\) is:
\[ \boxed{3} \]