For `(1)/(asqrt(x)+bsqrt(y))` the rationalising factor is a `asqrt(x)-bsqrt(y)` .
If `x=(1)/(3-2sqrt(2))` and `y=(1)/(3+2sqrt(2))` , then value of `xy^(2)+x^(2)y` is

Options

A. `4`
B. `12`
C. `6`
D. `9`

Answer

To solve the problem, we need to find the value of \( xy^2 + x^2y \) given that \( x = \frac{1}{3 - 2\sqrt{2}} \) and \( y = \frac{1}{3 + 2\sqrt{2}} \).

### Step-by-Step Solution:

1. **Rationalizing \( x \)**:
\[ x = \frac{1}{3 - 2\sqrt{2}} \]
To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator:
\[ x = \frac{1 \cdot (3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} \]
The denominator simplifies as follows:
\[ (3 - 2\sqrt{2})(3 + 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]
Therefore:
\[ x = 3 + 2\sqrt{2} \]

2. **Rationalizing \( y \)**:
\[ y = \frac{1}{3 + 2\sqrt{2}} \]
Similarly, we rationalize \( y \) by multiplying by the conjugate:
\[ y = \frac{1 \cdot (3 - 2\sqrt{2})}{(3 + 2\sqrt{2})(3 - 2\sqrt{2})} \]
The denominator simplifies to:
\[ (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]
Therefore:
\[ y = 3 - 2\sqrt{2} \]

3. **Calculating \( xy^2 + x^2y \)**:
We can factor the expression:
\[ xy^2 + x^2y = xy(y + x) \]
First, we find \( x + y \):
\[ x + y = (3 + 2\sqrt{2}) + (3 - 2\sqrt{2}) = 6 \]

4. **Calculating \( xy \)**:
Now we find \( xy \):
\[ xy = (3 + 2\sqrt{2})(3 - 2\sqrt{2}) = 3^2 - (2\sqrt{2})^2 = 9 - 8 = 1 \]

5. **Final Calculation**:
Now substituting back into the expression:
\[ xy^2 + x^2y = xy(y + x) = 1 \cdot 6 = 6 \]

Thus, the value of \( xy^2 + x^2y \) is \( \boxed{6} \).