Expression `((a+1/b)^(1/x)(a-1/b)^(1/x))/((b+1/a)^(1/x)(b-1/a)^(1/x))` is an integer if

Options

A. `a = 4, b = 2 , x = 4`
B. `a = 16 , b = 2 , x = 4`
C. `a = 64 , b =2 , x = 8`
D. `a = 16 , b = 4 , x = 4`

Answer

To solve the expression

\[ \frac{(a + \frac{1}{b})^{\frac{1}{x}} (a - \frac{1}{b})^{\frac{1}{x}}}{(b + \frac{1}{a})^{\frac{1}{x}} (b - \frac{1}{a})^{\frac{1}{x}}} \]

and determine when it is an integer, we can follow these steps:

### Step 1: Simplify the Expression
We can rewrite the expression using the property of exponents:

\[ \frac{(a + \frac{1}{b})(a - \frac{1}{b})^{\frac{1}{x}}}{(b + \frac{1}{a})(b - \frac{1}{a})^{\frac{1}{x}}} \]

This allows us to combine the powers:

\[ = \left(\frac{(a + \frac{1}{b})(a - \frac{1}{b})}{(b + \frac{1}{a})(b - \frac{1}{a})}\right)^{\frac{1}{x}} \]

### Step 2: Apply the Difference of Squares
Using the difference of squares, we can simplify the numerator and denominator:

\[ = \left(\frac{a^2 - \frac{1}{b^2}}{b^2 - \frac{1}{a^2}}\right)^{\frac{1}{x}} \]

### Step 3: Further Simplification
Now, we can rewrite the numerator and denominator:

\[ = \left(\frac{a^2b^2 - 1}{b^2a^2 - 1}\right)^{\frac{1}{x}} \]

### Step 4: Cancel Common Terms
Since the numerator and denominator are the same, we can simplify:

\[ = \left(\frac{a^2b^2 - 1}{b^2a^2 - 1}\right)^{\frac{1}{x}} = 1^{\frac{1}{x}} = 1 \]

### Step 5: Determine Conditions for Integer
The expression is an integer if the base of the exponent is an integer. Therefore, we need to check the values of \(a\), \(b\), and \(x\) such that:

\[ \frac{a^2b^2 - 1}{b^2a^2 - 1} \text{ is an integer} \]

### Step 6: Check Given Options
Now we can substitute the values from the options provided:

1. **Option A:** \(a = 4, b = 2, x = 4\)
\[ \frac{(4^2)(2^2) - 1}{(2^2)(4^2) - 1} = \frac{16 \cdot 4 - 1}{4 \cdot 16 - 1} = \frac{64 - 1}{64 - 1} = \frac{63}{63} = 1 \text{ (integer)} \]

2. **Option B:** \(a = 16, b = 2, x = 4\)
\[ \frac{(16^2)(2^2) - 1}{(2^2)(16^2) - 1} = \frac{256 \cdot 4 - 1}{4 \cdot 256 - 1} = \frac{1024 - 1}{1024 - 1} = \frac{1023}{1023} = 1 \text{ (integer)} \]

3. **Option C:** \(a = 64, b = 2, x = 8\)
\[ \frac{(64^2)(2^2) - 1}{(2^2)(64^2) - 1} = \frac{4096 \cdot 4 - 1}{4 \cdot 4096 - 1} = \frac{16384 - 1}{16384 - 1} = \frac{16383}{16383} = 1 \text{ (integer)} \]

4. **Option D:** \(a = 16, b = 4, x = 4\)
\[ \frac{(16^2)(4^2) - 1}{(4^2)(16^2) - 1} = \frac{256 \cdot 16 - 1}{16 \cdot 256 - 1} = \frac{4096 - 1}{4096 - 1} = \frac{4095}{4095} = 1 \text{ (integer)} \]

### Conclusion
The expression is an integer for all the given options.