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 $$ 16\left(m^{2}+n^{2}\right)^{2}+56\left(m^{2}+n^{2}\right)\left(3 m^{2}-2 n^{2}\right)+49\left(3 m^{2}-2 n^{2}\right)^{2} $$

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**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.
**Problem 6:** Given $$ a + \frac{1}{a} = 4 $$, find $$ a^4 + \frac{1}{a^4} $$. 1. Square both sides: $$ \left(a + \frac{1}{a}\right)^2 = 16 \implies a^2 + 2 + \frac{1}{a^2} = 16 \implies a^2 + \frac{1}{a^2} = 14 $$ 2. Square again: $$ \left(a^2 + \frac{1}{a^2}\right)^2 = 196 \implies a^4 + 2 + \frac{1}{a^4} = 196 \implies a^4 + \frac{1}{a^4} = 194 $$ **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. Let $$ A = m^2 + n^2 $$ and $$ B = 3m^2 - 2n^2 $$. The expression becomes: $$ 16A^2 + 56AB + 49B^2 = (4A + 7B)^2 $$ 2. Substitute $$ m = 6 $$ and $$ n = 7 $$: $$ m^2 = 36, \quad n^2 = 49 $$ $$ 4A + 7B = 4(36 + 49) + 7(3(36) - 2(49)) = 4(85) + 7(108 - 98) = 340 + 70 = 410 $$ 3. Square the result: $$ 410^2 = 168100 $$ **Answer:** The value of the expression is 168100.

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