Find the product of `((1)/(2)x^(2)-(1)/(3)y^(2))and((1)/(2)x^(2)+(1)/(3)y^(2))`

Options

A. `(1)/(9)y^(4)-(1)/(4)x^(4)`
B. `(1)/(4)x^(4)-(1)/(9)y^(4)`
C. `4x^(2)-9y^(4)`
D. `(1)/(y)x^(4)-(1)/(9)y^(9)`

Answer

To find the product of the expressions \(\left(\frac{1}{2}x^2 - \frac{1}{3}y^2\right)\) and \(\left(\frac{1}{2}x^2 + \frac{1}{3}y^2\right)\), we can use the difference of squares formula, which states that:

\[ (a - b)(a + b) = a^2 - b^2 \]

### Step-by-Step Solution:

1. **Identify \(A\) and \(B\)**:
- Let \(A = \frac{1}{2}x^2\)
- Let \(B = \frac{1}{3}y^2\)

2. **Apply the difference of squares formula**:
- The product can be rewritten as:
\[ (A - B)(A + B) = A^2 - B^2 \]

3. **Calculate \(A^2\)**:
- \(A^2 = \left(\frac{1}{2}x^2\right)^2 = \frac{1}{4}x^4\)

4. **Calculate \(B^2\)**:
- \(B^2 = \left(\frac{1}{3}y^2\right)^2 = \frac{1}{9}y^4\)

5. **Subtract \(B^2\) from \(A^2\)**:
- Now substitute \(A^2\) and \(B^2\) into the formula:
\[ A^2 - B^2 = \frac{1}{4}x^4 - \frac{1}{9}y^4 \]

6. **Final Result**:
- The final product is:
\[ \frac{1}{4}x^4 - \frac{1}{9}y^4 \]