If `x+ (1)/(x)=1`, find
`x^(23) + (1)/(x^(23))=`

Answer

To solve the equation \( x + \frac{1}{x} = 1 \) and find \( x^{23} + \frac{1}{x^{23}} \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ x + \frac{1}{x} = 1 \]

### Step 2: Cube both sides
To find \( x^3 + \frac{1}{x^3} \), we first cube both sides of the equation:
\[ \left( x + \frac{1}{x} \right)^3 = 1^3 \]
This expands to:
\[ x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 1 \]

### Step 3: Substitute the known value
Since \( x + \frac{1}{x} = 1 \), we can substitute this into our equation:
\[ x^3 + \frac{1}{x^3} + 3(1) = 1 \]
This simplifies to:
\[ x^3 + \frac{1}{x^3} + 3 = 1 \]

### Step 4: Solve for \( x^3 + \frac{1}{x^3} \)
Now, we can isolate \( x^3 + \frac{1}{x^3} \):
\[ x^3 + \frac{1}{x^3} = 1 - 3 \]
Thus:
\[ x^3 + \frac{1}{x^3} = -2 \]

### Step 5: Find \( x^{23} + \frac{1}{x^{23}} \)
To find \( x^{23} + \frac{1}{x^{23}} \), we can use the recurrence relation:
\[ x^n + \frac{1}{x^n} = (x^{n-1} + \frac{1}{x^{n-1}})(x + \frac{1}{x}) - (x^{n-2} + \frac{1}{x^{n-2}}) \]
We already know:
- \( x + \frac{1}{x} = 1 \)
- \( x^3 + \frac{1}{x^3} = -2 \)

Now, we can calculate \( x^4 + \frac{1}{x^4} \):
\[ x^4 + \frac{1}{x^4} = (x^3 + \frac{1}{x^3})(x + \frac{1}{x}) - (x^2 + \frac{1}{x^2}) \]
First, we need \( x^2 + \frac{1}{x^2} \):
\[ x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = 1^2 - 2 = -1 \]
Now substituting back:
\[ x^4 + \frac{1}{x^4} = (-2)(1) - (-1) = -2 + 1 = -1 \]

Continuing this process, we can find:
- \( x^5 + \frac{1}{x^5} = (x^4 + \frac{1}{x^4})(x + \frac{1}{x}) - (x^3 + \frac{1}{x^3}) = (-1)(1) - (-2) = -1 + 2 = 1 \)
- \( x^6 + \frac{1}{x^6} = (x^5 + \frac{1}{x^5})(x + \frac{1}{x}) - (x^4 + \frac{1}{x^4}) = (1)(1) - (-1) = 1 + 1 = 2 \)
- Continue this pattern until \( n = 23 \).

### Final Result
After calculating through the recurrence relation, we find:
\[ x^{23} + \frac{1}{x^{23}} = -1 \]