If `x = a cos^(2) theta sin theta` and `y = a sin^(2) theta cos theta `, then `( x^(2) + y^(2))^(3)` is

Options

A. `a^(2) x^(2) //y^(2)`
B. `a^(2) x^(2) y ^(2)`
C. `a^(2) ( x^(2) - y^(2))`
D. none

Answer

To solve the problem, we need to find the value of \((x^2 + y^2)^3\) given that \(x = a \cos^2 \theta \sin \theta\) and \(y = a \sin^2 \theta \cos \theta\).

### Step-by-Step Solution:

1. **Substitute the values of \(x\) and \(y\)**:
\[ x^2 + y^2 = (a \cos^2 \theta \sin \theta)^2 + (a \sin^2 \theta \cos \theta)^2 \]

2. **Expand the squares**:
\[ = a^2 \cos^4 \theta \sin^2 \theta + a^2 \sin^4 \theta \cos^2 \theta \]

3. **Factor out the common terms**:
\[ = a^2 \cos^2 \theta \sin^2 \theta (\cos^2 \theta + \sin^2 \theta) \]

4. **Use the Pythagorean identity**:
Since \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ = a^2 \cos^2 \theta \sin^2 \theta \cdot 1 \]
\[ = a^2 \cos^2 \theta \sin^2 \theta \]

5. **Now, we need to cube this expression**:
\[ (x^2 + y^2)^3 = (a^2 \cos^2 \theta \sin^2 \theta)^3 \]
\[ = a^6 \cos^6 \theta \sin^6 \theta \]

6. **Final expression**:
\[ = a^6 \cos^6 \theta \sin^6 \theta \]

### Final Answer:
\[ (x^2 + y^2)^3 = a^6 \cos^6 \theta \sin^6 \theta \]