If `b-c`, `bx-cy`, `bx^(2)-cy^(2)` (`b,c ne 0`) are in `G.P`, then the value of `((bx+cy)/(b+c))((bx-cy)/(b-c))` is

Options

A. `x^(2)`
B. `-x^(2)`
C. `2y^(2)`
D. `3y^(2)`

Explanation

To solve the problem, we need to find the value of the expression \(\frac{(bx + cy)}{(b + c)} \cdot \frac{(bx - cy)}{(b - c)}\) given that \(b - c\), \(bx - cy\), and \(bx^2 - cy^2\) are in geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the condition of G.P.**:
Since \(b - c\), \(bx - cy\), and \(bx^2 - cy^2\) are in G.P., we can use the property that if \(a\), \(b\), \(c\) are in G.P., then \(b^2 = ac\).
Here, we can set:
- \(a = b - c\)
- \(b = bx - cy\)
- \(c = bx^2 - cy^2\)

Thus, we have:
\[ (bx - cy)^2 = (b - c)(bx^2 - cy^2) \]

2. **Expanding both sides**:
Expanding the left side:
\[ (bx - cy)^2 = b^2x^2 - 2bcyx + c^2y^2 \]
Expanding the right side:
\[ (b - c)(bx^2 - cy^2) = b(bx^2) - b(cy^2) - c(bx^2) + c(cy^2) = b^2x^2 - bcy^2 - bcx^2 + c^2y^2 \]

3. **Setting the equations equal**:
Now we equate both expanded forms:
\[ b^2x^2 - 2bcyx + c^2y^2 = b^2x^2 - bcy^2 - bcx^2 + c^2y^2 \]

4. **Cancelling common terms**:
Cancel \(b^2x^2\) and \(c^2y^2\) from both sides:
\[ -2bcyx = -bcx^2 - bcy^2 \]

5. **Rearranging the equation**:
Rearranging gives:
\[ 2bcyx = bcy^2 + bcx^2 \]
Dividing through by \(bc\) (since \(b, c \neq 0\)):
\[ 2yx = y^2 + x^2 \]

6. **Rearranging to a standard form**:
Rearranging gives:
\[ y^2 - 2yx + x^2 = 0 \]
This can be factored as:
\[ (y - x)^2 = 0 \]
Thus, we find:
\[ y = x \]

7. **Substituting \(y = x\) into the expression**:
Now substituting \(y = x\) into the expression we need to evaluate:
\[ \frac{(bx + cy)}{(b + c)} \cdot \frac{(bx - cy)}{(b - c)} = \frac{(bx + cx)}{(b + c)} \cdot \frac{(bx - cx)}{(b - c)} \]
This simplifies to:
\[ \frac{x(b + c)}{(b + c)} \cdot \frac{x(b - c)}{(b - c)} \]
Cancelling out the terms gives:
\[ x^2 \]

### Final Answer:
Thus, the value of the expression is:
\[ \boxed{x^2} \]