If `alpha, beta, gamma` are such that `alpha +beta+gamma=2, alpha^(2)+beta^(2)+gamma^(2)=6, alpha^(3)+beta^(3)+gamma^(3)=8`, then `alpha^(4)+beta^(4)+gamma^(4)` is equal to:

Options

A. 10
B. 12
C. 18
D. none of these

Answer

To find the value of \( \alpha^4 + \beta^4 + \gamma^4 \) given the equations:

1. \( \alpha + \beta + \gamma = 2 \)
2. \( \alpha^2 + \beta^2 + \gamma^2 = 6 \)
3. \( \alpha^3 + \beta^3 + \gamma^3 = 8 \)

we can use the relationships between the symmetric sums of the roots.

### Step 1: Use the identity for the square of the sum of the roots

We know that:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the known values:
\[ 6 = 2^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
This simplifies to:
\[ 6 = 4 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Rearranging gives:
\[ 2(\alpha\beta + \beta\gamma + \gamma\alpha) = 4 - 6 = -2 \]
Thus:
\[ \alpha\beta + \beta\gamma + \gamma\alpha = -1 \]

### Step 2: Use the identity for the sum of cubes

Next, we use the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma \]
Substituting the known values:
\[ 8 = 2(6 - (-1)) + 3\alpha\beta\gamma \]
This simplifies to:
\[ 8 = 2(6 + 1) + 3\alpha\beta\gamma \]
\[ 8 = 2 \times 7 + 3\alpha\beta\gamma \]
\[ 8 = 14 + 3\alpha\beta\gamma \]
Rearranging gives:
\[ 3\alpha\beta\gamma = 8 - 14 = -6 \]
Thus:
\[ \alpha\beta\gamma = -2 \]

### Step 3: Use the identity for the fourth power

Now we can find \( \alpha^4 + \beta^4 + \gamma^4 \) using the identity:
\[ \alpha^4 + \beta^4 + \gamma^4 = (\alpha^2 + \beta^2 + \gamma^2)^2 - 2(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2) \]
First, we compute \( (\alpha^2 + \beta^2 + \gamma^2)^2 \):
\[ (\alpha^2 + \beta^2 + \gamma^2)^2 = 6^2 = 36 \]

Next, we need to find \( \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \):
Using the identity:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the known values:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (-1)^2 - 2(-2)(2) \]
\[ = 1 + 8 = 9 \]

### Step 4: Substitute back to find \( \alpha^4 + \beta^4 + \gamma^4 \)

Now we can substitute back:
\[ \alpha^4 + \beta^4 + \gamma^4 = 36 - 2 \times 9 = 36 - 18 = 18 \]

Thus, the final answer is:
\[ \alpha^4 + \beta^4 + \gamma^4 = 18 \]