If `xya= (3x+y)^2 - (3x−y)^2` , then the value of a is

Options

A. 12
B. 6
C. 0
D. 18

Answer

To solve the equation \( xya = (3x+y)^2 - (3x-y)^2 \), we can follow these steps:

### Step 1: Expand the squares
We start by expanding both squares in the expression:
\[ (3x+y)^2 = (3x)^2 + 2(3x)(y) + y^2 = 9x^2 + 6xy + y^2 \]
\[ (3x-y)^2 = (3x)^2 - 2(3x)(y) + y^2 = 9x^2 - 6xy + y^2 \]

### Step 2: Substitute the expanded forms
Now, we substitute these expansions back into the original equation:
\[ xya = (9x^2 + 6xy + y^2) - (9x^2 - 6xy + y^2) \]

### Step 3: Simplify the expression
Next, we simplify the right side:
\[ xya = (9x^2 + 6xy + y^2) - 9x^2 + 6xy - y^2 \]
This simplifies to:
\[ xya = 6xy + 6xy = 12xy \]

### Step 4: Solve for \( a \)
Now, we can express \( a \) in terms of \( x \) and \( y \):
\[ xya = 12xy \]
Dividing both sides by \( xy \) (assuming \( xy \neq 0 \)):
\[ a = 12 \]

Thus, the value of \( a \) is \( 12 \).

### Final Answer:
The value of \( a \) is \( 12 \).

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