Find the common roots of `x^12-1=0 and x^4+x^2+1=0`

Explanation

To find the common roots of the equations \(x^{12} - 1 = 0\) and \(x^4 + x^2 + 1 = 0\), we can follow these steps:

### Step 1: Solve \(x^{12} - 1 = 0\)
The equation \(x^{12} - 1 = 0\) can be factored as:
\[ x^{12} - 1 = (x^6 - 1)(x^6 + 1) \]
Next, we can further factor \(x^6 - 1\):
\[ x^6 - 1 = (x^3 - 1)(x^3 + 1) \]
And we can factor these further:
\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \]
\[ x^3 + 1 = (x + 1)(x^2 - x + 1) \]
Thus, we have:
\[ x^{12} - 1 = (x - 1)(x + 1)(x^2 + x + 1)(x^2 - x + 1)(x^6 + 1) \]

The roots of \(x^{12} - 1 = 0\) are the 12th roots of unity:
\[ x = e^{2\pi i k / 12}, \quad k = 0, 1, 2, \ldots, 11 \]
These roots are:
\[ 1, \omega, \omega^2, \omega^3, \omega^4, \omega^5, -1, -\omega, -\omega^2, -\omega^3, -\omega^4, -\omega^5 \]
where \(\omega = e^{\pi i / 6}\).

### Step 2: Solve \(x^4 + x^2 + 1 = 0\)
Let \(y = x^2\). Then, we can rewrite the equation as:
\[ y^2 + y + 1 = 0 \]
Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, the roots for \(y\) are:
\[ y_1 = \frac{-1 + i\sqrt{3}}{2}, \quad y_2 = \frac{-1 - i\sqrt{3}}{2} \]
Returning to \(x\), we find:
\[ x^2 = y_1 \implies x = \pm \sqrt{y_1}, \quad x^2 = y_2 \implies x = \pm \sqrt{y_2} \]

### Step 3: Find common roots
The roots of \(x^4 + x^2 + 1 = 0\) can be expressed in terms of the cube roots of unity. Specifically, we know:
\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{-2\pi i / 3} \]
Thus, the roots of \(x^4 + x^2 + 1 = 0\) can be expressed as:
\[ x = \pm \omega, \quad x = \pm \omega^2 \]

### Conclusion
The common roots of \(x^{12} - 1 = 0\) and \(x^4 + x^2 + 1 = 0\) are:
\[ x = \omega, \quad x = -\omega, \quad x = \omega^2, \quad x = -\omega^2 \]