If `alpha, beta , gamma, delta` are the roots of the equation `x^4+x^2+1=0` then the equation whose roots are `alpha^2, beta^2, gamma^2, delta^2` is

Options

A. `(x^(2) - x + 1)^(2) = 0`
B. `(x^(2) + x + 1)^(2) = 0`
C. `x^(4) - x^(2) + 1 = 0`
D. `x^(2) + x + 1 = 0`

Answer

To find the equation whose roots are \( \alpha^2, \beta^2, \gamma^2, \delta^2 \) given that \( \alpha, \beta, \gamma, \delta \) are the roots of the equation \( x^4 + x^2 + 1 = 0 \), we can follow these steps:

### Step 1: Substitute \( x^2 = t \)
We start with the equation:
\[ x^4 + x^2 + 1 = 0 \]
Let \( t = x^2 \). Then, we can rewrite the equation as:
\[ t^2 + t + 1 = 0 \]

### Step 2: Find the roots of the quadratic equation
To find the roots of \( t^2 + t + 1 = 0 \), we use the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = 1 \):
\[ t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} \]
This simplifies to:
\[ t = \frac{-1 \pm i\sqrt{3}}{2} \]
Let \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \).

### Step 3: Find the roots of \( x^2 \)
The roots \( t = \omega \) and \( t = \omega^2 \) correspond to:
\[ x^2 = \omega \quad \text{and} \quad x^2 = \omega^2 \]
The roots of \( x \) will be:
\[ x = \sqrt{\omega}, -\sqrt{\omega}, \sqrt{\omega^2}, -\sqrt{\omega^2} \]
Thus, the roots \( \alpha, \beta, \gamma, \delta \) are:
\[ \alpha = \sqrt{\omega}, \quad \beta = -\sqrt{\omega}, \quad \gamma = \sqrt{\omega^2}, \quad \delta = -\sqrt{\omega^2} \]

### Step 4: Calculate \( \alpha^2, \beta^2, \gamma^2, \delta^2 \)
Now we find \( \alpha^2, \beta^2, \gamma^2, \delta^2 \):
\[ \alpha^2 = \omega, \quad \beta^2 = \omega, \quad \gamma^2 = \omega^2, \quad \delta^2 = \omega^2 \]
Thus, the roots \( \alpha^2, \beta^2, \gamma^2, \delta^2 \) are \( \omega, \omega, \omega^2, \omega^2 \).

### Step 5: Form the new polynomial
The new polynomial can be formed using the roots:
\[ (x - \omega)^2 (x - \omega^2)^2 \]
Expanding this:
\[ = (x^2 - 2\omega x + \omega^2)(x^2 - 2\omega^2 x + \omega^4) \]
Using \( \omega^3 = 1 \) (since \( \omega \) is a cube root of unity):
\[ = (x^2 - 2\omega x + \omega^2)(x^2 - 2\omega^2 x + 1) \]

### Step 6: Final expansion
Now, we expand this product:
1. First, expand \( (x^2 - 2\omega x + \omega^2)(x^2) \).
2. Then expand \( (x^2 - 2\omega x + \omega^2)(-2\omega^2 x) \).
3. Finally, combine all terms.

The resulting polynomial will be:
\[ x^4 + 2x^2 + 1 = 0 \]

### Conclusion
Thus, the equation whose roots are \( \alpha^2, \beta^2, \gamma^2, \delta^2 \) is:
\[ x^4 + 2x^2 + 1 = 0 \]