The value of
`(3 + 2sqrt(2))^(-3) + (3 - 2sqrt(2))^(-3)` is

Options

A. 198
B. 180
C. 108
D. 189

Answer

To solve the expression \((3 + 2\sqrt{2})^{-3} + (3 - 2\sqrt{2})^{-3}\), we will follow these steps:

### Step 1: Rewrite the expression
We can rewrite the expression by moving the negative exponent to the denominator:
\[ (3 + 2\sqrt{2})^{-3} + (3 - 2\sqrt{2})^{-3} = \frac{1}{(3 + 2\sqrt{2})^3} + \frac{1}{(3 - 2\sqrt{2})^3} \]

### Step 2: Find a common denominator
The common denominator for the two fractions is \((3 + 2\sqrt{2})^3(3 - 2\sqrt{2})^3\). So we can write:
\[ \frac{(3 - 2\sqrt{2})^3 + (3 + 2\sqrt{2})^3}{(3 + 2\sqrt{2})^3(3 - 2\sqrt{2})^3} \]

### Step 3: Use the sum of cubes formula
The sum of cubes formula states that \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\). Here, let \(a = 3 + 2\sqrt{2}\) and \(b = 3 - 2\sqrt{2}\):
- \(a + b = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6\)
- \(ab = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1\)

Now we can find \(a^2 + b^2\):
\[ a^2 + b^2 = (a + b)^2 - 2ab = 6^2 - 2 \cdot 1 = 36 - 2 = 34 \]

### Step 4: Substitute into the formula
Now substituting back into the sum of cubes formula:
\[ (3 + 2\sqrt{2})^3 + (3 - 2\sqrt{2})^3 = (6)(34 - 1) = 6 \cdot 33 = 198 \]

### Step 5: Calculate the denominator
Next, we need to calculate the denominator:
\[ (3 + 2\sqrt{2})^3(3 - 2\sqrt{2})^3 = ((3 + 2\sqrt{2})(3 - 2\sqrt{2}))^3 = (1)^3 = 1 \]

### Step 6: Final result
Thus, we have:
\[ \frac{198}{1} = 198 \]

So, the final value of \((3 + 2\sqrt{2})^{-3} + (3 - 2\sqrt{2})^{-3}\) is:
\[ \boxed{198} \]

---