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 that \(\alpha, \beta, \gamma\) are the roots of the polynomial \(x^3 + 2x^2 + 1 = 0\).

### Step 1: Identify the coefficients of the polynomial
The polynomial can be expressed in the standard form \(x^3 + ax^2 + bx + c\), where:
- \(a = 2\)
- \(b = 0\)
- \(c = 1\)

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
- \(\alpha + \beta + \gamma = -a = -2\)
- \(\alpha\beta + \beta\gamma + \gamma\alpha = b = 0\)
- \(\alpha\beta\gamma = -c = -1\)

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

Thus, we need to compute:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 \]

### Step 4: Calculate \(a + b + c\)
From Vieta's, we have:
\[ \alpha \beta + \beta \gamma + \alpha \gamma = 0 \]

### Step 5: Calculate \(a^2 + b^2 + c^2\)
Using the square of the sum:
\[ a^2 + b^2 + c^2 = (\alpha \beta)^2 + (\beta \gamma)^2 + (\alpha \gamma)^2 \]
We can express this in terms of \(\alpha, \beta, \gamma\):
\[ = (\alpha \beta)^2 + (\beta \gamma)^2 + (\alpha \gamma)^2 = \alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 \]

### Step 6: Substitute values
Using the identity:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 = 3\alpha \beta \gamma \cdot 0 + (\alpha \beta + \beta \gamma + \alpha \gamma)((\alpha \beta)^2 + (\beta \gamma)^2 + (\alpha \gamma)^2 - (\alpha \beta)(\beta \gamma) - (\beta \gamma)(\alpha \gamma) - (\alpha \gamma)(\alpha \beta)) \]
This simplifies to:
\[ = 0 + 0 = 0 \]

### Step 7: Calculate \(\alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2\)
Using:
\[ \alpha^2 \beta^2 + \beta^2 \gamma^2 + \alpha^2 \gamma^2 = (\alpha \beta \gamma)^2 = (-1)^2 = 1 \]

### Step 8: Final Calculation
Thus, we find:
\[ (\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3 = 3 \cdot 1 = 3 \]

### Final Answer
The value of \((\alpha \beta)^3 + (\beta \gamma)^3 + (\alpha \gamma)^3\) is **3**.

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