`(x^3+1/x^3)+(x^2+1/x^2)-6(x+1/x)-7=0`

Options

A. `(3 + sqrt(5))/(2)`
B. `(-3 - sqrt(5))/(2)`
C. `(3 - sqrt(5))/(2)`
D. `(-3 + sqrt(5))/(2)`

Explanation

To solve the equation

\[ (x^3 + \frac{1}{x^3}) + (x^2 + \frac{1}{x^2}) - 6(x + \frac{1}{x}) - 7 = 0, \]

we will use the substitution \( t = x + \frac{1}{x} \).

### Step 1: Express \( x^2 + \frac{1}{x^2} \) and \( x^3 + \frac{1}{x^3} \) in terms of \( t \)

Using the identity for squares:

\[ x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2 = t^2 - 2. \]

Using the identity for cubes:

\[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x})^3 - 3(x + \frac{1}{x}) = t^3 - 3t. \]

### Step 2: Substitute these expressions into the original equation

Substituting \( x^2 + \frac{1}{x^2} \) and \( x^3 + \frac{1}{x^3} \) into the equation gives:

\[ (t^3 - 3t) + (t^2 - 2) - 6t - 7 = 0. \]

### Step 3: Simplify the equation

Combining like terms:

\[ t^3 + t^2 - 9t - 9 = 0. \]

### Step 4: Factor the cubic equation

We can use the Rational Root Theorem or trial and error to find a root. Testing \( t = -1 \):

\[ (-1)^3 + (-1)^2 - 9(-1) - 9 = -1 + 1 + 9 - 9 = 0. \]

Thus, \( t + 1 \) is a factor. We can divide the cubic polynomial by \( t + 1 \):

Using synthetic division or polynomial long division, we find:

\[ t^3 + t^2 - 9t - 9 = (t + 1)(t^2 - 9). \]

### Step 5: Factor further

The quadratic \( t^2 - 9 \) can be factored as:

\[ (t - 3)(t + 3). \]

Thus, we have:

\[ (t + 1)(t - 3)(t + 3) = 0. \]

### Step 6: Solve for \( t \)

Setting each factor to zero gives:

1. \( t + 1 = 0 \) → \( t = -1 \)
2. \( t - 3 = 0 \) → \( t = 3 \)
3. \( t + 3 = 0 \) → \( t = -3 \)

### Step 7: Solve for \( x \)

Recall \( t = x + \frac{1}{x} \).

1. For \( t = -1 \):
\[ x + \frac{1}{x} = -1 \implies x^2 + x + 1 = 0. \]
The discriminant is negative, so no real roots.

2. For \( t = 3 \):
\[ x + \frac{1}{x} = 3 \implies x^2 - 3x + 1 = 0. \]
Using the quadratic formula:
\[ x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{3 \pm \sqrt{5}}{2}. \]

3. For \( t = -3 \):
\[ x + \frac{1}{x} = -3 \implies x^2 + 3x + 1 = 0. \]
Again using the quadratic formula:
\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2}. \]

### Final Solutions

The solutions for \( x \) are:

1. \( x = \frac{3 + \sqrt{5}}{2} \)
2. \( x = \frac{3 - \sqrt{5}}{2} \)
3. \( x = \frac{-3 + \sqrt{5}}{2} \)
4. \( x = \frac{-3 - \sqrt{5}}{2} \)