If `x^2 -3x + 1 =0`, and xy = 1 then find the value of `x^3-y^3 +3xy`

Options

A. 0
B. 2
C. 3
D. None of these

Answer

To solve the equation \( x^2 - 3x + 1 = 0 \) and find the value of \( x^3 - y^3 + 3xy \) given that \( xy = 1 \), we will follow these steps:

### Step 1: Solve for \( x \)
We start with the quadratic equation:
\[ x^2 - 3x + 1 = 0 \]
We can use the quadratic formula, which is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -3 \), and \( c = 1 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot 1 = 9 - 4 = 5 \]
Now substituting into the quadratic formula:
\[ x = \frac{3 \pm \sqrt{5}}{2} \]

### Step 2: Find \( y \)
Given \( xy = 1 \), we can express \( y \) in terms of \( x \):
\[ y = \frac{1}{x} \]

### Step 3: Calculate \( x^3 - y^3 + 3xy \)
We can use the identity for the difference of cubes:
\[ x^3 - y^3 = (x - y)(x^2 + xy + y^2) \]
We also know \( xy = 1 \), so we can express \( y \) in terms of \( x \).

Calculating \( x - y \):
\[ x - y = x - \frac{1}{x} = \frac{x^2 - 1}{x} \]

Calculating \( x^2 + xy + y^2 \):
\[ y^2 = \left(\frac{1}{x}\right)^2 = \frac{1}{x^2} \]
Thus,
\[ x^2 + xy + y^2 = x^2 + 1 + \frac{1}{x^2} \]

### Step 4: Find \( x^2 + \frac{1}{x^2} \)
To find \( x^2 + \frac{1}{x^2} \), we can use the identity:
\[ x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 \]
We already know:
\[ x + \frac{1}{x} = 3 \quad \text{(since } x + \frac{1}{x} \text{ can be derived from the original equation)} \]
Thus,
\[ x^2 + \frac{1}{x^2} = 3^2 - 2 = 9 - 2 = 7 \]
So,
\[ x^2 + xy + y^2 = 7 + 1 = 8 \]

### Step 5: Substitute back into the expression
Now we can substitute back into the expression for \( x^3 - y^3 + 3xy \):
\[ x^3 - y^3 + 3xy = (x - y)(x^2 + xy + y^2) + 3xy \]
Substituting the values we found:
\[ = \left(\frac{x^2 - 1}{x}\right)(8) + 3 \]
We need to calculate \( x^2 - 1 \):
\[ x^2 = \frac{3 \pm \sqrt{5}}{2}^2 = \frac{9 \pm 6\sqrt{5} + 5}{4} = \frac{14 \pm 6\sqrt{5}}{4} \]
Thus,
\[ x^2 - 1 = \frac{14 \pm 6\sqrt{5}}{4} - 1 = \frac{10 \pm 6\sqrt{5}}{4} = \frac{5 \pm 3\sqrt{5}}{2} \]

### Final Calculation
Now substituting back:
\[ = \left(\frac{\frac{5 \pm 3\sqrt{5}}{2}}{x}\right)(8) + 3 \]
This will yield a complicated expression, but we can simplify it further to find the final numerical value.

### Conclusion
After evaluating the above expression, we find that the value of \( x^3 - y^3 + 3xy \) is \( 8\sqrt{5} + 3 \).