The equation `x ^(4) -2x ^(3)-3x^2 + 4x -1=0` has four distinct real roots `x _(1), x _(2), x _(3), x_(4)` such that `x _(1) lt x _(2) lt x _(3)lt x _(4)` and product of two roots is unity, then : `x _(2)^(3) + x _(4)^(3)=`

Options

A. `(2+5 sqrt5)/(8)`
B. `-4`
C. `(27 sqrt5 + 5)/(4)`
D. 18

Answer

To solve the equation \( x^4 - 2x^3 - 3x^2 + 4x - 1 = 0 \) and find \( x_2^3 + x_4^3 \) given that the product of two roots is unity, we can follow these steps:

### Step 1: Factor the Polynomial
We start with the polynomial:
\[ x^4 - 2x^3 - 3x^2 + 4x - 1 = 0 \]
We can try to factor it into two quadratic equations. Let's assume:
\[ (x^2 + ax + b)(x^2 + cx + d) = 0 \]
By expanding and comparing coefficients, we can find suitable values for \( a, b, c, \) and \( d \).

### Step 2: Set Up the Quadratic Equations
After some trials, we can factor the polynomial as:
\[ (x^2 - 3x + 1)(x^2 + x - 1) = 0 \]

### Step 3: Solve Each Quadratic Equation
Now we solve each quadratic equation separately.

1. For \( x^2 - 3x + 1 = 0 \):
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2} \]
This gives us the roots:
\[ x_3 = \frac{3 + \sqrt{5}}{2}, \quad x_2 = \frac{3 - \sqrt{5}}{2} \]

2. For \( x^2 + x - 1 = 0 \):
Again using the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
This gives us the roots:
\[ x_1 = \frac{-1 - \sqrt{5}}{2}, \quad x_4 = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Order the Roots
Now we have:
\[ x_1 < x_2 < x_3 < x_4 \]
where:
- \( x_1 = \frac{-1 - \sqrt{5}}{2} \)
- \( x_2 = \frac{3 - \sqrt{5}}{2} \)
- \( x_3 = \frac{3 + \sqrt{5}}{2} \)
- \( x_4 = \frac{-1 + \sqrt{5}}{2} \)

### Step 5: Calculate \( x_2^3 + x_4^3 \)
We need to find \( x_2^3 + x_4^3 \). We can use the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
Let \( a = x_2 \) and \( b = x_4 \).

1. Calculate \( x_2 + x_4 \):
\[ x_2 + x_4 = \frac{3 - \sqrt{5}}{2} + \frac{-1 + \sqrt{5}}{2} = \frac{2}{2} = 1 \]

2. Calculate \( x_2 x_4 \):
\[ x_2 x_4 = \left(\frac{3 - \sqrt{5}}{2}\right)\left(\frac{-1 + \sqrt{5}}{2}\right) = \frac{(3)(-1) + (3)(\sqrt{5}) - (\sqrt{5})(1) + (\sqrt{5})(\sqrt{5})}{4} = \frac{-3 + 2\sqrt{5} + 5}{4} = \frac{2 + 2\sqrt{5}}{4} = \frac{1 + \sqrt{5}}{2} \]

3. Calculate \( x_2^2 + x_4^2 \):
\[ x_2^2 + x_4^2 = (x_2 + x_4)^2 - 2x_2 x_4 = 1^2 - 2 \cdot \frac{1 + \sqrt{5}}{2} = 1 - (1 + \sqrt{5}) = -\sqrt{5} \]

4. Now substitute into the identity:
\[ x_2^3 + x_4^3 = (x_2 + x_4)((x_2^2 + x_4^2) - x_2 x_4) = 1 \left(-\sqrt{5} - \frac{1 + \sqrt{5}}{2}\right) \]
Simplifying:
\[ = -\sqrt{5} - \frac{1 + \sqrt{5}}{2} = -\frac{2\sqrt{5}}{2} - \frac{1 + \sqrt{5}}{2} = -\frac{2\sqrt{5} + 1 + \sqrt{5}}{2} = -\frac{3\sqrt{5} + 1}{2} \]

### Final Answer
Thus, the value of \( x_2^3 + x_4^3 \) is:
\[ -\frac{3\sqrt{5} + 1}{2} \]