If `x^2 - 3x + 1 = 0`, then the value of `(x^6 + x^4 + x^2 + 1)/(x^3)` will be

Options

A. 18
B. 15
C. 21
D. 30

Answer

To solve the equation \( x^2 - 3x + 1 = 0 \) and find the value of \( \frac{x^6 + x^4 + x^2 + 1}{x^3} \), we can follow these steps:

### Step 1: Solve the quadratic equation
The given equation is:
\[ x^2 - 3x + 1 = 0 \]
We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -3, c = 1 \). Plugging in these values:
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]

### Step 2: Find \( x + \frac{1}{x} \)
To find \( x + \frac{1}{x} \), we first divide the original equation by \( x \):
\[ \frac{x^2}{x} - \frac{3x}{x} + \frac{1}{x} = 0 \implies x - 3 + \frac{1}{x} = 0 \implies x + \frac{1}{x} = 3 \]

### Step 3: Find \( x^2 + \frac{1}{x^2} \)
Using the identity:
\[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \]
We can rearrange it to find \( x^2 + \frac{1}{x^2} \):
\[ x^2 + \frac{1}{x^2} = \left(x + \frac{1}{x}\right)^2 - 2 = 3^2 - 2 = 9 - 2 = 7 \]

### Step 4: Find \( x^3 + \frac{1}{x^3} \)
Using the identity:
\[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 + \frac{1}{x^2}\right) - \left(x + \frac{1}{x}\right) \]
Substituting the values we found:
\[ x^3 + \frac{1}{x^3} = 3 \cdot 7 - 3 = 21 - 3 = 18 \]

### Step 5: Calculate \( x^6 + x^4 + x^2 + 1 \)
We can express \( x^6 + x^4 + x^2 + 1 \) using the values we have:
\[ x^6 + x^4 + x^2 + 1 = (x^6 + 1) + (x^4 + x^2) = (x^3 + \frac{1}{x^3})^2 - 2 + (x^2 + \frac{1}{x^2}) = 18^2 - 2 + 7 \]
Calculating \( 18^2 \):
\[ 18^2 = 324 \]
Thus,
\[ x^6 + x^4 + x^2 + 1 = 324 - 2 + 7 = 329 \]

### Step 6: Divide by \( x^3 \)
Now, we need to find:
\[ \frac{x^6 + x^4 + x^2 + 1}{x^3} = \frac{329}{x^3} \]
Using \( x^3 + \frac{1}{x^3} = 18 \), we can express \( x^3 \) in terms of \( x + \frac{1}{x} \):
\[ \frac{329}{x^3} = \frac{329}{\frac{18}{x^3}} = \frac{329}{18} \]

### Final Answer
Thus, the value of \( \frac{x^6 + x^4 + x^2 + 1}{x^3} \) is:
\[ \frac{329}{18} \]