If `(x^(2) + 1)/(x)= 3(1)/(3) and x gt 1`, find
`x^(3)- (1)/(x^(3))`

Answer

To solve the problem, we need to find the value of \( x^3 - \frac{1}{x^3} \) given that \( \frac{x^2 + 1}{x} = \frac{10}{3} \) and \( x > 1 \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \frac{x^2 + 1}{x} = \frac{10}{3} \]
This can be rewritten as:
\[ x^2 + 1 = \frac{10}{3} x \]

2. **Rearranging the equation:**
\[ x^2 - \frac{10}{3} x + 1 = 0 \]

3. **Multiply through by 3 to eliminate the fraction:**
\[ 3x^2 - 10x + 3 = 0 \]

4. **Use the quadratic formula to find \( x \):**
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3 \), \( b = -10 \), and \( c = 3 \):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ x = \frac{10 \pm \sqrt{100 - 36}}{6} \]
\[ x = \frac{10 \pm \sqrt{64}}{6} \]
\[ x = \frac{10 \pm 8}{6} \]

5. **Calculate the possible values of \( x \):**
\[ x = \frac{18}{6} = 3 \quad \text{or} \quad x = \frac{2}{6} = \frac{1}{3} \]
Since we are given \( x > 1 \), we take \( x = 3 \).

6. **Now, find \( x^2 + \frac{1}{x^2} \):**
We know:
\[ x^2 + \frac{1}{x^2} = \left( \frac{x^2 + 1}{x} \right)^2 - 2 \]
First, calculate \( \frac{x^2 + 1}{x} = \frac{10}{3} \):
\[ x^2 + \frac{1}{x^2} = \left( \frac{10}{3} \right)^2 - 2 \]
\[ = \frac{100}{9} - 2 = \frac{100}{9} - \frac{18}{9} = \frac{82}{9} \]

7. **Next, find \( x - \frac{1}{x} \):**
\[ x - \frac{1}{x} = \sqrt{x^2 + \frac{1}{x^2} - 2} \]
\[ = \sqrt{\frac{82}{9} - 2} = \sqrt{\frac{82}{9} - \frac{18}{9}} = \sqrt{\frac{64}{9}} = \frac{8}{3} \]

8. **Now, use the identity to find \( x^3 - \frac{1}{x^3} \):**
The identity is:
\[ x^3 - \frac{1}{x^3} = \left( x - \frac{1}{x} \right) \left( x^2 + \frac{1}{x^2} + 1 \right) \]
Substitute the values:
\[ = \left( \frac{8}{3} \right) \left( \frac{82}{9} + 1 \right) \]
\[ = \left( \frac{8}{3} \right) \left( \frac{82}{9} + \frac{9}{9} \right) = \left( \frac{8}{3} \right) \left( \frac{91}{9} \right) \]
\[ = \frac{728}{27} \]

9. **Final result:**
\[ x^3 - \frac{1}{x^3} = \frac{728}{27} \]