All the roots of the equation `x^(11) - x^(6) - x^(5) + 1 = 0` lie on a circle of radius __________.

Answer

To find the radius of the circle on which all the roots of the equation \(x^{11} - x^{6} - x^{5} + 1 = 0\) lie, we can follow these steps:

### Step 1: Factor the equation
We start with the equation:
\[ x^{11} - x^{6} - x^{5} + 1 = 0 \]
We can factor out \(x^6\) from the first two terms and \(-1\) from the last two terms:
\[ x^6(x^5 - 1) - 1(x^5 - 1) = 0 \]
This can be rewritten as:
\[ (x^6 - 1)(x^5 - 1) = 0 \]

### Step 2: Solve the factored equations
Now we have two separate equations to solve:
1. \(x^6 - 1 = 0\)
2. \(x^5 - 1 = 0\)

### Step 3: Find the roots of \(x^6 - 1 = 0\)
The roots of \(x^6 - 1 = 0\) are the sixth roots of unity. These can be expressed as:
\[ x = e^{i\frac{2\pi k}{6}} \quad \text{for } k = 0, 1, 2, 3, 4, 5 \]
This gives us 6 roots, all lying on the unit circle (radius = 1).

### Step 4: Find the roots of \(x^5 - 1 = 0\)
The roots of \(x^5 - 1 = 0\) are the fifth roots of unity. These can be expressed as:
\[ x = e^{i\frac{2\pi m}{5}} \quad \text{for } m = 0, 1, 2, 3, 4 \]
This gives us 5 roots, also lying on the unit circle (radius = 1).

### Step 5: Conclusion
Since both sets of roots (from \(x^6 - 1\) and \(x^5 - 1\)) lie on the unit circle, we conclude that all the roots of the original equation lie on a circle of radius:
\[ \text{Radius} = 1 \]

### Final Answer
All the roots of the equation \(x^{11} - x^{6} - x^{5} + 1 = 0\) lie on a circle of radius **1**.

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