If `((x)/(y))=((z)/(w))`, then what is `(xy+zw)^(2)` equal to ?

Options

A. `(x^(2)+z^(2))(y^(2)+w^(2))`
B. `x^(2)y^(2)+z^(2)w^(2)`
C. `x^(2)w^(2)+y^(2)z^(2)`
D. `(x^(2)+w^(2))(y^(2)+z^(2))`

Answer

To solve the problem step by step, we start with the given equation and manipulate it to find the value of \((xy + zw)^2\).

### Step 1: Start with the given equation
We are given:
\[ \frac{x}{y} = \frac{z}{w} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ x \cdot w = z \cdot y \]

### Step 3: Rearrange the equation
Rearranging this equation, we can write:
\[ xw - zy = 0 \]

### Step 4: Square both sides
Now, we square both sides of the equation:
\[ (xw - zy)^2 = 0^2 \]
This simplifies to:
\[ (xw - zy)^2 = 0 \]

### Step 5: Apply the square of a binomial formula
Using the formula \((a - b)^2 = a^2 - 2ab + b^2\), we have:
\[ (xw)^2 - 2(xw)(zy) + (zy)^2 = 0 \]
This expands to:
\[ x^2w^2 - 2xwyz + z^2y^2 = 0 \]

### Step 6: Rearranging the equation
This can be rearranged as:
\[ x^2w^2 + z^2y^2 = 2xwyz \]

### Step 7: Find \((xy + zw)^2\)
Now, we can express \((xy + zw)^2\) using the formula \((a + b)^2 = a^2 + 2ab + b^2\):
\[ (xy + zw)^2 = (xy)^2 + 2(xy)(zw) + (zw)^2 \]

### Step 8: Substitute values
Substituting \(xy\) and \(zw\) into the equation:
\[ = x^2y^2 + 2xyzw + z^2w^2 \]

### Step 9: Combine the results
Now, we can relate this back to our earlier equation. We know from our previous steps that:
\[ x^2w^2 + z^2y^2 = 2xwyz \]
Thus, we can express \((xy + zw)^2\) as:
\[ = x^2y^2 + z^2w^2 + 2xyzw \]

### Final Result
Thus, we conclude that:
\[ (xy + zw)^2 = (xy + zw)^2 \]
This means that the expression simplifies to:
\[ = (xy + zw)^2 = (x^2y^2 + z^2w^2 + 2xyzw) \]