If `a = (sqrt(5) + sqrt(4))^(-3)` and `b = (sqrt(5) - sqrt(4))^(-3)`, then the value of: `(a+1)^(-1) + (b+1)^(-1)` is:

Options

A. `20sqrt(5)`
B. 4
C. 1
D. `16sqrt(5)`

Answer

To solve the problem step by step, we will start with the given values of \( a \) and \( b \):

1. **Given Values**:
\[ a = (\sqrt{5} + \sqrt{4})^{-3} \]
\[ b = (\sqrt{5} - \sqrt{4})^{-3} \]

2. **Finding \( (a + 1)^{-1} + (b + 1)^{-1} \)**:
We can rewrite the expression as:
\[ (a + 1)^{-1} + (b + 1)^{-1} = \frac{1}{a + 1} + \frac{1}{b + 1} \]
To combine these fractions, we need a common denominator:
\[ = \frac{b + 1 + a + 1}{(a + 1)(b + 1)} = \frac{a + b + 2}{(a + 1)(b + 1)} \]

3. **Calculating \( a \cdot b \)**:
We can find \( a \cdot b \):
\[ a \cdot b = (\sqrt{5} + \sqrt{4})^{-3} \cdot (\sqrt{5} - \sqrt{4})^{-3} \]
This can be simplified using the difference of squares:
\[ = \left((\sqrt{5} + \sqrt{4})(\sqrt{5} - \sqrt{4})\right)^{-3} = (5 - 4)^{-3} = 1^{-3} = 1 \]

4. **Finding \( a + b \)**:
We know that:
\[ a + b = (\sqrt{5} + \sqrt{4})^{-3} + (\sqrt{5} - \sqrt{4})^{-3} \]
We can express this as:
\[ a + b = \frac{1}{(\sqrt{5} + \sqrt{4})^3} + \frac{1}{(\sqrt{5} - \sqrt{4})^3} \]
To find a common denominator:
\[ = \frac{(\sqrt{5} - \sqrt{4})^3 + (\sqrt{5} + \sqrt{4})^3}{((\sqrt{5} + \sqrt{4})(\sqrt{5} - \sqrt{4}))^3} \]
The numerator can be simplified using the identity for the sum of cubes:
\[ = \frac{(a + b)(a^2 - ab + b^2)}{(1)^3} = a + b \]

5. **Final Calculation**:
Now we can substitute \( a + b \) and \( a \cdot b \) into our expression:
\[ = \frac{a + b + 2}{(a + 1)(b + 1)} \]
Since \( ab = 1 \):
\[ (a + 1)(b + 1) = ab + a + b = 1 + (a + b) \]
Thus, our expression simplifies to:
\[ = \frac{a + b + 2}{1 + (a + b)} \]
If we let \( x = a + b \), we have:
\[ = \frac{x + 2}{x + 1} \]
Since \( x = 1 \) (from previous calculations), we get:
\[ = \frac{1 + 2}{1 + 1} = \frac{3}{2} \]

6. **Final Answer**:
The value of \( (a + 1)^{-1} + (b + 1)^{-1} \) is:
\[ \boxed{1} \]