If ` alpha and beta ` are the complex cube roots of unity, then ` alpha^(4) + beta^(4) + alpha ^(-1) beta^(-1)`

Options

A. 1
B. 2
C. 3
D. 0

Answer

To solve the problem, we need to evaluate the expression \( \alpha^4 + \beta^4 + \alpha^{-1} \beta^{-1} \), where \( \alpha \) and \( \beta \) are the complex cube roots of unity.

### Step 1: Identify the cube roots of unity
The complex cube roots of unity are given by:
\[ 1, \omega, \omega^2 \]
where \( \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \) and \( \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i \).

### Step 2: Assign values to \( \alpha \) and \( \beta \)
Let \( \alpha = \omega \) and \( \beta = \omega^2 \).

### Step 3: Calculate \( \alpha^4 \) and \( \beta^4 \)
Using the property of powers of roots of unity, we know:
\[ \omega^3 = 1 \quad \text{and} \quad \omega^4 = \omega \quad \text{and} \quad (\omega^2)^4 = \omega^2 \]
Thus, we have:
\[ \alpha^4 = \omega^4 = \omega \]
\[ \beta^4 = (\omega^2)^4 = \omega^2 \]

### Step 4: Calculate \( \alpha^{-1} \) and \( \beta^{-1} \)
The inverses of \( \alpha \) and \( \beta \) are:
\[ \alpha^{-1} = \frac{1}{\omega} = \omega^2 \quad \text{and} \quad \beta^{-1} = \frac{1}{\omega^2} = \omega \]
Thus,
\[ \alpha^{-1} \beta^{-1} = \omega^2 \cdot \omega = \omega^3 = 1 \]

### Step 5: Combine all parts to find the total
Now we substitute back into the original expression:
\[ \alpha^4 + \beta^4 + \alpha^{-1} \beta^{-1} = \omega + \omega^2 + 1 \]

### Step 6: Use the property of cube roots of unity
From the property of the cube roots of unity, we know:
\[ 1 + \omega + \omega^2 = 0 \]
Thus,
\[ \omega + \omega^2 + 1 = 0 \]

### Final Result
The value of \( \alpha^4 + \beta^4 + \alpha^{-1} \beta^{-1} \) is:
\[ \boxed{0} \]