Which of the following is the best approximation to `((2^(3)-1) (3^(3)-1)……..(1000^(3)-1))/((2^(3)+1)(3^(3)+1)…..(1000^(3) +1))`

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{(2^3 - 1)(3^3 - 1) \ldots (1000^3 - 1)}{(2^3 + 1)(3^3 + 1) \ldots (1000^3 + 1)} \]

### Step 1: Rewrite the expression
We can express the numerator and denominator in a more manageable form. The numerator can be written as:

\[ \prod_{r=2}^{1000} (r^3 - 1) = \prod_{r=2}^{1000} (r - 1)(r^2 + r + 1) \]

And the denominator as:

\[ \prod_{r=2}^{1000} (r^3 + 1) = \prod_{r=2}^{1000} (r + 1)(r^2 - r + 1) \]

Thus, we can rewrite the entire expression as:

\[ \frac{\prod_{r=2}^{1000} (r - 1)(r^2 + r + 1)}{\prod_{r=2}^{1000} (r + 1)(r^2 - r + 1)} \]

### Step 2: Simplify the expression
Now we can separate the products:

\[ = \frac{\prod_{r=2}^{1000} (r - 1)}{\prod_{r=2}^{1000} (r + 1)} \cdot \frac{\prod_{r=2}^{1000} (r^2 + r + 1)}{\prod_{r=2}^{1000} (r^2 - r + 1)} \]

### Step 3: Evaluate the first product
The first product simplifies as follows:

\[ \frac{(1)(2)(3)\ldots(999)}{(3)(4)(5)\ldots(1001)} = \frac{1}{1001} \cdot \frac{999!}{1000!} = \frac{1}{1001} \]

### Step 4: Evaluate the second product
For the second product, we need to analyze:

\[ \frac{r^2 + r + 1}{r^2 - r + 1} \]

For large \( r \), this approaches:

\[ \frac{r^2 + r + 1}{r^2 - r + 1} \approx 1 + \frac{2r}{r^2 - r + 1} \approx 1 + \frac{2}{r} \]

Thus, the product from \( r = 2 \) to \( r = 1000 \) can be approximated using logarithms:

\[ \prod_{r=2}^{1000} \left(1 + \frac{2}{r}\right) \approx e^{\sum_{r=2}^{1000} \frac{2}{r}} \approx e^{2 \ln(1000)} = 1000^2 \]

### Step 5: Combine results
Now we combine the results:

\[ \frac{1}{1001} \cdot 1000^2 = \frac{1000000}{1001} \]

### Step 6: Approximate the final result
To approximate \( \frac{1000000}{1001} \):

\[ \frac{1000000}{1001} \approx 998.001998 \approx 998 \]

### Conclusion
Thus, the best approximation to the original expression is:

\[ \boxed{998} \]