If `x^4 + 1/(x^4) = 119 and x > 1`. Then find the positive value of `x^3 - 1/(x^3)` .

Options

A. 25
B. 27
C. 36
D. 49

Answer

To solve the problem step by step, we start with the given equation:

1. **Given Equation**:
\[ x^4 + \frac{1}{x^4} = 119 \]

2. **Add 2 to Both Sides**:
We can rewrite the left side as a square:
\[ x^4 + 2 + \frac{1}{x^4} = 119 + 2 \]
This simplifies to:
\[ (x^2 + \frac{1}{x^2})^2 = 121 \]

3. **Take the Square Root**:
Taking the square root of both sides gives us:
\[ x^2 + \frac{1}{x^2} = 11 \]

4. **Use the Identity for Cubes**:
We need to find \( x^3 - \frac{1}{x^3} \). We can use the identity:
\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x})(x^2 - 1 + \frac{1}{x^2}) \]
First, we need to find \( x + \frac{1}{x} \).

5. **Find \( x^2 - \frac{1}{x^2} \)**:
We can use the identity:
\[ (x - \frac{1}{x})^2 = x^2 - 2 + \frac{1}{x^2} \]
Rearranging gives us:
\[ x^2 + \frac{1}{x^2} = (x - \frac{1}{x})^2 + 2 \]
Since we know \( x^2 + \frac{1}{x^2} = 11 \), we can set:
\[ (x - \frac{1}{x})^2 + 2 = 11 \]
Thus:
\[ (x - \frac{1}{x})^2 = 9 \]
Taking the square root gives:
\[ x - \frac{1}{x} = 3 \quad (\text{since } x > 1) \]

6. **Find \( x + \frac{1}{x} \)**:
We can find \( x + \frac{1}{x} \) using the identity:
\[ (x + \frac{1}{x})^2 = (x - \frac{1}{x})^2 + 4 \]
Substituting the value we found:
\[ (x + \frac{1}{x})^2 = 9 + 4 = 13 \]
Taking the square root gives:
\[ x + \frac{1}{x} = \sqrt{13} \]

7. **Calculate \( x^3 - \frac{1}{x^3} \)**:
Now we can substitute back into our identity:
\[ x^3 - \frac{1}{x^3} = (x + \frac{1}{x})((x^2 + \frac{1}{x^2}) - 1) \]
Substituting the known values:
\[ x^3 - \frac{1}{x^3} = \sqrt{13} \cdot (11 - 1) = \sqrt{13} \cdot 10 = 10\sqrt{13} \]

However, we need to find the value of \( x^3 - \frac{1}{x^3} \) directly from the previous steps.

8. **Final Calculation**:
Using the earlier derived formula:
\[ x^3 - \frac{1}{x^3} = (x - \frac{1}{x})^3 + 3(x - \frac{1}{x}) \]
We have:
\[ (3)^3 + 3 \cdot 3 = 27 + 9 = 36 \]

Thus, the positive value of \( x^3 - \frac{1}{x^3} \) is:
\[ \boxed{36} \]