If `x+ (1)/(x)= -1` find
`x^(27) + (1)/(x^(27)) =` ?

Answer

To solve the equation \( x + \frac{1}{x} = -1 \) and find \( x^{27} + \frac{1}{x^{27}} \), we can follow these steps:

### Step 1: Cube both sides of the equation
Starting with the equation:
\[ x + \frac{1}{x} = -1 \]
We will cube both sides:
\[ \left( x + \frac{1}{x} \right)^3 = (-1)^3 \]

### Step 2: Apply the identity for cubing a binomial
Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \), where \( a = x \) and \( b = \frac{1}{x} \):
\[ x^3 + \frac{1}{x^3} + 3 \left( x \cdot \frac{1}{x} \right) \left( x + \frac{1}{x} \right) = -1 \]
Since \( x \cdot \frac{1}{x} = 1 \), we can simplify:
\[ x^3 + \frac{1}{x^3} + 3 \cdot 1 \cdot (-1) = -1 \]
This simplifies to:
\[ x^3 + \frac{1}{x^3} - 3 = -1 \]

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

### Step 4: Find \( x^9 + \frac{1}{x^9} \)
Next, we need to find \( x^9 + \frac{1}{x^9} \). We can use the identity again:
\[ \left( x^3 + \frac{1}{x^3} \right)^3 = x^9 + \frac{1}{x^9} + 3 \left( x^3 \cdot \frac{1}{x^3} \right) \left( x^3 + \frac{1}{x^3} \right) \]
Substituting \( x^3 + \frac{1}{x^3} = 2 \):
\[ 2^3 = x^9 + \frac{1}{x^9} + 3 \cdot 1 \cdot 2 \]
This simplifies to:
\[ 8 = x^9 + \frac{1}{x^9} + 6 \]
Rearranging gives:
\[ x^9 + \frac{1}{x^9} = 8 - 6 = 2 \]

### Step 5: Find \( x^{27} + \frac{1}{x^{27}} \)
Now we can find \( x^{27} + \frac{1}{x^{27}} \) using the same identity:
\[ \left( x^9 + \frac{1}{x^9} \right)^3 = x^{27} + \frac{1}{x^{27}} + 3 \left( x^9 \cdot \frac{1}{x^9} \right) \left( x^9 + \frac{1}{x^9} \right) \]
Substituting \( x^9 + \frac{1}{x^9} = 2 \):
\[ 2^3 = x^{27} + \frac{1}{x^{27}} + 3 \cdot 1 \cdot 2 \]
This simplifies to:
\[ 8 = x^{27} + \frac{1}{x^{27}} + 6 \]
Rearranging gives:
\[ x^{27} + \frac{1}{x^{27}} = 8 - 6 = 2 \]

### Final Answer
Thus, the final result is:
\[ \boxed{2} \]