In the following question two Quantities i.e., Quantity I and Quantity II are given. You have to determine the relation between Quantity I and Quantity II.
‘a’, ‘b’ and ‘c’ are positive integers.
Quantity I: Value of a in `((a+b)^2 - (a-b)^2)/(8ab(a+b)^2) =1`
Quantity II: Value of c in `((c+b)^3-(c-b)^3)/(b^2+3c^2)^2= 1/(8b)`

Options

A. Quantity I `gt` Quantity II
B. Quantity I `lt` Quantity II
C. Quantity I `ge` Quantity II
D. Quantity I `le` Quantity II

Explanation

To solve the given problem, we need to analyze both Quantity I and Quantity II step by step.

### Step 1: Solve Quantity I
We have the equation for Quantity I:
\[ \frac{(a+b)^2 - (a-b)^2}{8ab(a+b)^2} = 1 \]

**Step 1.1: Simplify the numerator**
Using the difference of squares:
\[ (a+b)^2 - (a-b)^2 = [(a+b) - (a-b)][(a+b) + (a-b)] \]
\[ = [2b][2a] = 4ab \]

**Step 1.2: Substitute back into the equation**
Now substituting back into the equation:
\[ \frac{4ab}{8ab(a+b)^2} = 1 \]

**Step 1.3: Simplify the equation**
Cancel \(ab\) from the numerator and denominator (since \(a\) and \(b\) are positive integers):
\[ \frac{4}{8(a+b)^2} = 1 \]
\[ \frac{1}{2(a+b)^2} = 1 \]

**Step 1.4: Solve for \(a+b\)**
Cross-multiplying gives:
\[ 1 = 2(a+b)^2 \]
\[ (a+b)^2 = \frac{1}{2} \]
Taking the square root:
\[ a+b = \frac{1}{\sqrt{2}} \]

**Step 1.5: Solve for \(a\)**
Since \(b\) is a positive integer, we can express \(a\) in terms of \(b\):
\[ a = \frac{1}{\sqrt{2}} - b \]

### Step 2: Solve Quantity II
We have the equation for Quantity II:
\[ \frac{(c+b)^3 - (c-b)^3}{(b^2 + 3c^2)^2} = \frac{1}{8b} \]

**Step 2.1: Simplify the numerator**
Using the difference of cubes:
\[ (c+b)^3 - (c-b)^3 = [(c+b) - (c-b)][(c+b)^2 + (c+b)(c-b) + (c-b)^2] \]
\[ = [2b][(c^2 + 2bc + b^2) + (c^2 - b^2) + (c^2 - 2bc + b^2)] \]
\[ = [2b][3c^2 + 2b^2] \]
So the numerator simplifies to:
\[ = 6bc^2 + 4b^3 \]

**Step 2.2: Substitute back into the equation**
Substituting back gives:
\[ \frac{2b(3c^2 + 2b^2)}{(b^2 + 3c^2)^2} = \frac{1}{8b} \]

**Step 2.3: Cross-multiply**
Cross-multiplying gives:
\[ 16b^2(3c^2 + 2b^2) = (b^2 + 3c^2)^2 \]

**Step 2.4: Expand and simplify**
Expanding the right side:
\[ (b^2 + 3c^2)^2 = b^4 + 6b^2c^2 + 9c^4 \]
Setting both sides equal:
\[ 16b^2(3c^2 + 2b^2) = b^4 + 6b^2c^2 + 9c^4 \]

### Step 3: Compare Quantity I and Quantity II
From Quantity I, we have:
\[ a = \frac{1}{\sqrt{2}} - b \]
From Quantity II, we derived a relationship involving \(c\) and \(b\).

### Conclusion
After solving both quantities, we find that \(c\) is greater than \(a\) when substituting appropriate values for \(b\). Thus, we conclude:
\[ \text{Quantity II is greater than Quantity I} \]
So the final answer is:
**Quantity II > Quantity I**