Find the root of the equation `|x-1|^(x^(2)+x-2)=1`

Answer

To solve the equation \( |x-1|^{(x^2+x-2)} = 1 \), we can analyze the conditions under which the expression equals 1.

### Step 1: Identify conditions for the equation to hold true
The expression \( a^b = 1 \) can be true under the following conditions:
1. \( a = 1 \) (for any \( b \))
2. \( a = -1 \) (for even \( b \))
3. \( b = 0 \) (for any \( a \neq 0 \))

### Step 2: Analyze the first condition \( |x-1| = 1 \)
We start with the first condition:
\[ |x-1| = 1 \]
This gives us two cases:
1. \( x - 1 = 1 \) → \( x = 2 \)
2. \( x - 1 = -1 \) → \( x = 0 \)

### Step 3: Analyze the second condition \( |x-1| = -1 \)
Since the absolute value cannot be negative, we do not have any valid solutions from this condition.

### Step 4: Analyze the third condition \( x^2 + x - 2 = 0 \)
Next, we check when the exponent is zero:
\[ x^2 + x - 2 = 0 \]
Factoring the quadratic:
\[ (x-1)(x+2) = 0 \]
This gives us:
1. \( x - 1 = 0 \) → \( x = 1 \)
2. \( x + 2 = 0 \) → \( x = -2 \)

### Step 5: Compile all solutions
From the analysis, we have the following potential solutions:
- From \( |x-1| = 1 \): \( x = 2 \) and \( x = 0 \)
- From \( x^2 + x - 2 = 0 \): \( x = 1 \) and \( x = -2 \)

### Step 6: Check for validity of solutions
1. **For \( x = 2 \)**:
\[ |2-1|^{(2^2 + 2 - 2)} = 1^2 = 1 \quad \text{(valid)} \]
2. **For \( x = 0 \)**:
\[ |0-1|^{(0^2 + 0 - 2)} = 1^{-2} = 1 \quad \text{(valid)} \]
3. **For \( x = 1 \)**:
\[ |1-1|^{(1^2 + 1 - 2)} = 0^0 \quad \text{(indeterminate, not valid)} \]
4. **For \( x = -2 \)**:
\[ |-2-1|^{((-2)^2 + (-2) - 2)} = 3^0 = 1 \quad \text{(valid)} \]

### Final Solutions
The valid solutions to the equation \( |x-1|^{(x^2+x-2)} = 1 \) are:
\[ \boxed{2, 0, -2} \]