If x = 3 + `sqrt(8) , y = 3 - sqrt(8)`, find the value of `x^(-3) + y^(-3)`

Options

A. 195
B. 199
C. 198
D. 201

Answer

To find the value of \( x^{-3} + y^{-3} \), where \( x = 3 + \sqrt{8} \) and \( y = 3 - \sqrt{8} \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ x^{-3} + y^{-3} = \frac{1}{x^3} + \frac{1}{y^3} \]
This can be combined into a single fraction:
\[ x^{-3} + y^{-3} = \frac{y^3 + x^3}{x^3 y^3} \]

### Step 2: Find \( x^3 + y^3 \)
Using the identity for the sum of cubes:
\[ x^3 + y^3 = (x + y)(x^2 - xy + y^2) \]
We need to find \( x + y \) and \( xy \).

### Step 3: Calculate \( x + y \)
\[ x + y = (3 + \sqrt{8}) + (3 - \sqrt{8}) = 6 \]

### Step 4: Calculate \( xy \)
Using the difference of squares:
\[ xy = (3 + \sqrt{8})(3 - \sqrt{8}) = 3^2 - (\sqrt{8})^2 = 9 - 8 = 1 \]

### Step 5: Calculate \( x^2 + y^2 \)
We can find \( x^2 + y^2 \) using the identity:
\[ x^2 + y^2 = (x + y)^2 - 2xy \]
Substituting the values we found:
\[ x^2 + y^2 = 6^2 - 2 \cdot 1 = 36 - 2 = 34 \]

### Step 6: Substitute back into the sum of cubes formula
Now we can substitute back into the sum of cubes formula:
\[ x^3 + y^3 = (x + y)((x^2 + y^2) - xy) = 6(34 - 1) = 6 \cdot 33 = 198 \]

### Step 7: Calculate \( x^3 y^3 \)
Since \( xy = 1 \):
\[ x^3 y^3 = (xy)^3 = 1^3 = 1 \]

### Step 8: Combine the results
Now we can substitute \( x^3 + y^3 \) and \( x^3 y^3 \) back into our expression:
\[ x^{-3} + y^{-3} = \frac{x^3 + y^3}{x^3 y^3} = \frac{198}{1} = 198 \]

### Final Answer
Thus, the value of \( x^{-3} + y^{-3} \) is:
\[ \boxed{198} \]