If `x+ (1)/(x)=1`, find
`x^(10) + (1)/(x^(10))=`

Answer

To solve the equation \( x + \frac{1}{x} = 1 \) and find \( x^{10} + \frac{1}{x^{10}} \), 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 can use the identity:
\[ \left( a + b \right)^3 = a^3 + b^3 + 3ab(a + b) \]
Letting \( a = x \) and \( b = \frac{1}{x} \), we cube both sides:
\[ \left( x + \frac{1}{x} \right)^3 = 1^3 \]
This gives us:
\[ x^3 + \frac{1}{x^3} + 3 \cdot x \cdot \frac{1}{x} \cdot (x + \frac{1}{x}) = 1 \]
Since \( x \cdot \frac{1}{x} = 1 \), we simplify:
\[ x^3 + \frac{1}{x^3} + 3 \cdot 1 \cdot 1 = 1 \]
Thus:
\[ x^3 + \frac{1}{x^3} + 3 = 1 \]

### Step 3: Solve for \( x^3 + \frac{1}{x^3} \)
Rearranging gives:
\[ x^3 + \frac{1}{x^3} = 1 - 3 = -2 \]

### Step 4: Find \( x^6 + \frac{1}{x^6} \)
Next, we use the identity again:
\[ \left( x^3 + \frac{1}{x^3} \right)^2 = x^6 + \frac{1}{x^6} + 2 \]
Substituting \( x^3 + \frac{1}{x^3} = -2 \):
\[ (-2)^2 = x^6 + \frac{1}{x^6} + 2 \]
This simplifies to:
\[ 4 = x^6 + \frac{1}{x^6} + 2 \]
Thus:
\[ x^6 + \frac{1}{x^6} = 4 - 2 = 2 \]

### Step 5: Find \( x^9 + \frac{1}{x^9} \)
Now we find \( x^9 + \frac{1}{x^9} \) using:
\[ x^9 + \frac{1}{x^9} = (x^6 + \frac{1}{x^6})(x^3 + \frac{1}{x^3}) - (x^3 + \frac{1}{x^3}) \]
Substituting the known values:
\[ x^9 + \frac{1}{x^9} = (2)(-2) - (-2) \]
This simplifies to:
\[ x^9 + \frac{1}{x^9} = -4 + 2 = -2 \]

### Step 6: Find \( x^{10} + \frac{1}{x^{10}} \)
Finally, we find \( x^{10} + \frac{1}{x^{10}} \) using:
\[ x^{10} + \frac{1}{x^{10}} = (x^9 + \frac{1}{x^9})(x + \frac{1}{x}) - (x^8 + \frac{1}{x^8}) \]
We already have \( x + \frac{1}{x} = 1 \) and need \( x^8 + \frac{1}{x^8} \):
\[ x^8 + \frac{1}{x^8} = (x^6 + \frac{1}{x^6})(x^2 + \frac{1}{x^2}) - (x^4 + \frac{1}{x^4}) \]
We need \( x^2 + \frac{1}{x^2} \):
\[ \left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \]
So:
\[ 1^2 = x^2 + \frac{1}{x^2} + 2 \implies 1 = x^2 + \frac{1}{x^2} + 2 \implies x^2 + \frac{1}{x^2} = -1 \]
Now substituting back:
\[ x^4 + \frac{1}{x^4} = (x^2 + \frac{1}{x^2})^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1 \]
Now substituting into \( x^8 + \frac{1}{x^8} \):
\[ x^8 + \frac{1}{x^8} = (2)(-1) - (-1) = -2 + 1 = -1 \]
Finally substituting into \( x^{10} + \frac{1}{x^{10}} \):
\[ x^{10} + \frac{1}{x^{10}} = (-2)(1) - (-1) = -2 + 1 = -1 \]

### Final Answer
Thus, the value of \( x^{10} + \frac{1}{x^{10}} \) is:
\[ \boxed{-1} \]