6. If \( a+\frac{1}{a}=4 \), what is the value of \( a^{4}+\frac{1}{a^{4}} \) ? 7. If \( m=6, n=7 \), find the value of \( 16\left(m^{2}+n^{2}\right)^{2}+56\left(m^{2}+n^{2}\right)\left(3 m^{2}-2 n^{2}\right)+49\left(3 m^{2}-2 n^{2}\right)^{2} \)

**Problem 6:** Given \( a + \frac{1}{a} = 4 \), we want to find \( a^4 + \frac{1}{a^4} \). 1. **Find \( a^2 + \frac{1}{a^2} \):** Square both sides of the given equation: \[ \left(a + \frac{1}{a}\right)^2 = 4^2 \implies a^2 + 2\left(a \cdot \frac{1}{a}\right) + \frac{1}{a^2} = 16 \] Simplifying: \[ a^2 + \frac{1}{a^2} + 2 = 16 \implies a^2 + \frac{1}{a^2} = 14 \] 2. **Find \( a^4 + \frac{1}{a^4} \):** Square \( a^2 + \frac{1}{a^2} \): \[ \left(a^2 + \frac{1}{a^2}\right)^2 = 14^2 \implies a^4 + 2\left(a^2 \cdot \frac{1}{a^2}\right) + \frac{1}{a^4} = 196 \] Simplifying: \[ a^4 + \frac{1}{a^4} + 2 = 196 \implies a^4 + \frac{1}{a^4} = 194 \] **Answer:** \( a^4 + \frac{1}{a^4} = 194 \) --- **Problem 7:** Given \( m = 6 \) and \( n = 7 \), compute: \[ 16(m^2 + n^2)^2 + 56(m^2 + n^2)(3m^2 - 2n^2) + 49(3m^2 - 2n^2)^2 \] 1. **Simplify the Expression:** Let \( A = m^2 + n^2 \) and \( B = 3m^2 - 2n^2 \). The expression becomes: \[ 16A^2 + 56AB + 49B^2 \] Notice that this is a perfect square: \[ 16A^2 + 56AB + 49B^2 = (4A + 7B)^2 \] 2. **Express in Terms of \( m \) and \( n \):** \[ 4A + 7B = 4(m^2 + n^2) + 7(3m^2 - 2n^2) = 4m^2 + 4n^2 + 21m^2 - 14n^2 = 25m^2 - 10n^2 \] 3. **Substitute \( m = 6 \) and \( n = 7 \):** \[ m^2 = 36, \quad n^2 = 49 \] \[ 25m^2 - 10n^2 = 25(36) - 10(49) = 900 - 490 = 410 \] Therefore, the expression becomes: \[ (410)^2 = 168100 \] **Answer:** The value of the expression is 168100.