If ` x^(4) + (1)/( x^(4))` = 119 then the values of ` x^(3) + (1)/( x^(3))` are

Options

A. `pm 10 sqrt(13)`
B. ` pm sqrt(13)`
C. ` pm 16 sqrt(13)`
D. ` pm 13 sqrt(13)`

Answer

To solve the equation \( x^4 + \frac{1}{x^4} = 119 \) and find the values of \( x^3 + \frac{1}{x^3} \), we can follow these steps:

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ x^4 + \frac{1}{x^4} = 119 \]

2. **Relate \( x^4 + \frac{1}{x^4} \) to \( x^2 + \frac{1}{x^2} \):**
We know that:
\[ x^4 + \frac{1}{x^4} = \left( x^2 + \frac{1}{x^2} \right)^2 - 2 \]
Therefore, we can rewrite the equation as:
\[ \left( x^2 + \frac{1}{x^2} \right)^2 - 2 = 119 \]

3. **Solve for \( x^2 + \frac{1}{x^2} \):**
Adding 2 to both sides:
\[ \left( x^2 + \frac{1}{x^2} \right)^2 = 121 \]
Taking the square root of both sides:
\[ x^2 + \frac{1}{x^2} = \pm 11 \]
Since \( x^2 + \frac{1}{x^2} \) must be positive, we take:
\[ x^2 + \frac{1}{x^2} = 11 \]

4. **Relate \( x^2 + \frac{1}{x^2} \) to \( x + \frac{1}{x} \):**
We know that:
\[ x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 \]
Thus, we can write:
\[ \left( x + \frac{1}{x} \right)^2 - 2 = 11 \]
Adding 2 to both sides gives:
\[ \left( x + \frac{1}{x} \right)^2 = 13 \]

5. **Solve for \( x + \frac{1}{x} \):**
Taking the square root:
\[ x + \frac{1}{x} = \pm \sqrt{13} \]
Again, since \( x + \frac{1}{x} \) must be positive, we take:
\[ x + \frac{1}{x} = \sqrt{13} \]

6. **Find \( x^3 + \frac{1}{x^3} \):**
We use the identity:
\[ x^3 + \frac{1}{x^3} = \left( x + \frac{1}{x} \right)^3 - 3 \left( x + \frac{1}{x} \right) \]
Substituting \( x + \frac{1}{x} = \sqrt{13} \):
\[ x^3 + \frac{1}{x^3} = \left( \sqrt{13} \right)^3 - 3 \sqrt{13} \]
Calculating \( \left( \sqrt{13} \right)^3 \):
\[ \left( \sqrt{13} \right)^3 = 13\sqrt{13} \]
Thus:
\[ x^3 + \frac{1}{x^3} = 13\sqrt{13} - 3\sqrt{13} = 10\sqrt{13} \]

### Final Answer:
The values of \( x^3 + \frac{1}{x^3} \) are:
\[ \boxed{10\sqrt{13}} \]