`(a + (1)/(a)) ^(2) = 3 ` then the value of `a ^(18) + a ^(12) + a ^(6) + 1` is

Options

A. 3
B. 2
C. 0
D. 1

Answer

To solve the equation \((a + \frac{1}{a})^2 = 3\) and find the value of \(a^{18} + a^{12} + a^{6} + 1\), we can follow these steps:

### Step 1: Simplify the given equation
Start with the equation:
\[ (a + \frac{1}{a})^2 = 3 \]
Taking the square root of both sides, we get:
\[ a + \frac{1}{a} = \sqrt{3} \quad \text{or} \quad a + \frac{1}{a} = -\sqrt{3} \]

### Step 2: Find \(a^6\)
From the identity \(a + \frac{1}{a} = x\), we can derive a relationship for \(a^n\). For \(x = \sqrt{3}\), we can use the formula:
\[ a^2 + \frac{1}{a^2} = (a + \frac{1}{a})^2 - 2 = 3 - 2 = 1 \]
Next, we find \(a^4 + \frac{1}{a^4}\):
\[ a^4 + \frac{1}{a^4} = (a^2 + \frac{1}{a^2})^2 - 2 = 1^2 - 2 = -1 \]
Now, we can find \(a^6\):
\[ a^6 + \frac{1}{a^6} = (a^4 + \frac{1}{a^4})(a^2 + \frac{1}{a^2}) - (a^2 + \frac{1}{a^2}) = (-1)(1) - 1 = -1 - 1 = -2 \]
Thus, we have:
\[ a^6 + \frac{1}{a^6} = -2 \implies a^6 = -1 \quad \text{(since } a^6 \text{ and } \frac{1}{a^6} \text{ are inverses)} \]

### Step 3: Calculate \(a^{18} + a^{12} + a^{6} + 1\)
Now we can substitute \(a^6 = -1\) into the expression:
\[ a^{18} + a^{12} + a^{6} + 1 = (a^6)^3 + (a^6)^2 + a^6 + 1 \]
Substituting \(a^6 = -1\):
\[ = (-1)^3 + (-1)^2 + (-1) + 1 \]
Calculating each term:
\[ = -1 + 1 - 1 + 1 = 0 \]

### Final Answer
Thus, the value of \(a^{18} + a^{12} + a^{6} + 1\) is:
\[ \boxed{0} \]