If `(x^(2) - x - 1)^(x^(2) - 7x + 12) = 1` then

Options

A. Positive integral solutions are `4`
B. Negative integral solutions are `2`
C. non positive integral solution are `2`
D. Total integral solutions are `6`

Explanation

To solve the equation \((x^2 - x - 1)^{(x^2 - 7x + 12)} = 1\), we will explore the three cases where the expression can equal 1.

### Step 1: Identify Cases
The equation \(a^b = 1\) can be satisfied under the following conditions:
1. \(b = 0\) (where \(a^0 = 1\) for any \(a\) except \(a = 0\))
2. \(a = 1\) (where \(1^b = 1\) for any \(b\))
3. \(a = -1\) and \(b\) is an even integer (where \((-1)^{\text{even}} = 1\))

### Step 2: Case 1 - \(b = 0\)
Set \(b = 0\):
\[ x^2 - 7x + 12 = 0 \]
Factoring the quadratic:
\[ (x - 3)(x - 4) = 0 \]
Thus, the solutions are:
\[ x = 3 \quad \text{and} \quad x = 4 \]

### Step 3: Case 2 - \(a = 1\)
Set \(a = 1\):
\[ x^2 - x - 1 = 1 \]
Rearranging gives:
\[ x^2 - x - 2 = 0 \]
Factoring:
\[ (x - 2)(x + 1) = 0 \]
Thus, the solutions are:
\[ x = 2 \quad \text{and} \quad x = -1 \]

### Step 4: Case 3 - \(a = -1\) and \(b\) is even
Set \(a = -1\):
\[ x^2 - x - 1 = -1 \]
Rearranging gives:
\[ x^2 - x = 0 \]
Factoring:
\[ x(x - 1) = 0 \]
Thus, the solutions are:
\[ x = 0 \quad \text{and} \quad x = 1 \]
Now, we need to check if \(b\) is even for these values of \(x\).

1. For \(x = 0\):
\[ b = 0^2 - 7(0) + 12 = 12 \quad (\text{even}) \]
2. For \(x = 1\):
\[ b = 1^2 - 7(1) + 12 = 6 \quad (\text{even}) \]

Both values satisfy the condition that \(b\) is even.

### Step 5: Collect All Solutions
From all cases, we have the solutions:
- From Case 1: \(x = 3, 4\)
- From Case 2: \(x = 2, -1\)
- From Case 3: \(x = 0, 1\)

Thus, the complete solution set is:
\[ x = 3, 4, 2, -1, 0, 1 \]

### Final Answer
The total number of solutions is 6.

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