f \( a-\frac{1}{a}=m \), show that \( a^{4}+\frac{1}{a^{4}}=m^{4}+4 m^{2}+2 \)

To show that \( a^{4} + \frac{1}{a^{4}} = m^{4} + 4m^{2} + 2 \) given \( a - \frac{1}{a} = m \), we can follow these steps: 1. **Start with the given equation:** \[ a - \frac{1}{a} = m \] 2. **Square both sides to find \( a^2 + \frac{1}{a^2} \):** \[ \left(a - \frac{1}{a}\right)^2 = m^2 \] \[ a^2 - 2\left(a \cdot \frac{1}{a}\right) + \frac{1}{a^2} = m^2 \] \[ a^2 + \frac{1}{a^2} - 2 = m^2 \] \[ a^2 + \frac{1}{a^2} = m^2 + 2 \] 3. **Square \( a^2 + \frac{1}{a^2} \) to find \( a^4 + \frac{1}{a^4} \):** \[ \left(a^2 + \frac{1}{a^2}\right)^2 = (m^2 + 2)^2 \] \[ a^4 + 2\left(a^2 \cdot \frac{1}{a^2}\right) + \frac{1}{a^4} = m^4 + 4m^2 + 4 \] \[ a^4 + \frac{1}{a^4} + 2 = m^4 + 4m^2 + 4 \] \[ a^4 + \frac{1}{a^4} = m^4 + 4m^2 + 2 \] Thus, we've shown that: \[ a^{4} + \frac{1}{a^{4}} = m^{4} + 4m^{2} + 2 \] **Answer:** After expanding, a⁴ + 1/a⁴ is m⁴ plus four m² plus two. Thus, a⁴ + 1/a⁴ = m⁴ + 4m² + 2