To solve the equation \( x^2 + \frac{1}{x^2} = 1 \) and find the value of \( x^{48} + x^{42} + x^{36} + x^{30} + x^{24} + x^{18} + x^{12} + x^{6} + 1 \), we will follow these steps:
### Step 1: Solve for \( x + \frac{1}{x} \)
Given:
\[
x^2 + \frac{1}{x^2} = 1
\]
We can rewrite this as:
\[
x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2
\]
So, we have:
\[
\left( x + \frac{1}{x} \right)^2 - 2 = 1
\]
Adding 2 to both sides:
\[
\left( x + \frac{1}{x} \right)^2 = 3
\]
Taking the square root:
\[
x + \frac{1}{x} = \sqrt{3} \quad \text{or} \quad x + \frac{1}{x} = -\sqrt{3}
\]
### Step 2: Find \( x^3 + \frac{1}{x^3} \)
Using the identity:
\[
x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3\left( x + \frac{1}{x} \right)
\]
Substituting \( x + \frac{1}{x} = \sqrt{3} \):
\[
x^3 + \frac{1}{x^3} = \left( \sqrt{3} \right)^3 - 3\sqrt{3} = 3\sqrt{3} - 3\sqrt{3} = 0
\]
### Step 3: Establish a recurrence relation
Let \( a_n = x^n + \frac{1}{x^n} \). We know:
- \( a_0 = 2 \)
- \( a_1 = \sqrt{3} \)
- \( a_2 = 1 \)
- \( a_3 = 0 \)
Using the recurrence relation:
\[
a_n = \left( x + \frac{1}{x} \right) a_{n-1} - a_{n-2}
\]
We can compute further values:
- For \( n = 4 \):
\[
a_4 = \sqrt{3} \cdot 0 - 1 = -1
\]
- For \( n = 5 \):
\[
a_5 = \sqrt{3} \cdot (-1) - 0 = -\sqrt{3}
\]
- For \( n = 6 \):
\[
a_6 = \sqrt{3} \cdot (-\sqrt{3}) - (-1) = -3 + 1 = -2
\]
- For \( n = 7 \):
\[
a_7 = \sqrt{3} \cdot (-2) - (-\sqrt{3}) = -2\sqrt{3} + \sqrt{3} = -\sqrt{3}
\]
- For \( n = 8 \):
\[
a_8 = \sqrt{3} \cdot (-\sqrt{3}) - (-2) = -3 + 2 = -1
\]
### Step 4: Identify the pattern
Continuing this way, we can observe that the values of \( a_n \) repeat every 6 terms:
- \( a_0 = 2 \)
- \( a_1 = \sqrt{3} \)
- \( a_2 = 1 \)
- \( a_3 = 0 \)
- \( a_4 = -1 \)
- \( a_5 = -\sqrt{3} \)
- \( a_6 = -2 \) (and then it repeats)
### Step 5: Calculate the required sum
Now, we need to find:
\[
x^{48} + x^{42} + x^{36} + x^{30} + x^{24} + x^{18} + x^{12} + x^{6} + 1
\]
Since \( 48, 42, 36, 30, 24, 18, 12, 6 \) all correspond to \( a_0, a_6, a_0, a_6, a_0, a_6, a_0, a_6 \) respectively, we can substitute:
\[
= 2 + (-2) + 2 + (-2) + 2 + (-2) + 2 + (-2) + 1
\]
Grouping the terms:
\[
(2 - 2) + (2 - 2) + (2 - 2) + (2 - 2) + 1 = 0 + 0 + 0 + 0 + 1 = 1
\]
### Final Answer
Thus, the value of \( x^{48} + x^{42} + x^{36} + x^{30} + x^{24} + x^{18} + x^{12} + x^{6} + 1 \) is:
\[
\boxed{1}
\]
