Question

If \( \alpha^2 - \frac{1}{\alpha^2} = 2 \), find \( \left( \alpha + \frac{1}{\alpha} \right)^{16} \).

Hint:

Remember to square both sides when dealing with expressions like \( \alpha^2 - \frac{1}{\alpha^2} \).

Answer 1

Step 1: Rearrange the given equation.
We are given \( \alpha^2 - \frac{1}{\alpha^2} = 2 \). We can express this as: \[ \left( \alpha + \frac{1}{\alpha} \right)^2 - 2 = 2 \] Simplifying: \[ \left( \alpha + \frac{1}{\alpha} \right)^2 = 4 \]
Step 2: Solve for \( \alpha + \frac{1}{\alpha} \).
Taking the square root of both sides: \[ \alpha + \frac{1}{\alpha} = 2 \quad \text{or} \quad \alpha + \frac{1}{\alpha} = -2 \]
Step 3: Compute \( \left( \alpha + \frac{1}{\alpha} \right)^{16} \).
Since we need to find \( \left( \alpha + \frac{1}{\alpha} \right)^{16} \), we have two cases: - If \( \alpha + \frac{1}{\alpha} = 2 \), then: \[ \left( \alpha + \frac{1}{\alpha} \right)^{16} = 2^{16} \] - If \( \alpha + \frac{1}{\alpha} = -2 \), then: \[ \left( \alpha + \frac{1}{\alpha} \right)^{16} = (-2)^{16} = 2^{16} \] Thus, in both cases: \[ \left( \alpha + \frac{1}{\alpha} \right)^{16} = 2^{16} \]