If `a=(sqrt(3)-sqrt(2))/(sqrt(3)+sqrt(2))` and `b=(sqrt(3)+sqrt(2))/(sqrt(3)-sqrt(2))` , then the value of `(a^(2))/(b)+(b^(2))/(a)` is :

Options

A. A) `1030`
B. B) `1025`
C. C) `970`
D. D) `930`

Answer

To solve the problem, we need to find the value of \(\frac{a^2}{b} + \frac{b^2}{a}\) given:

\[ a = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} \]
\[ b = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \]

### Step 1: Calculate \(a + b\)
First, we find \(a + b\):

\[ a + b = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} + \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \]

To add these fractions, we need a common denominator:

\[ = \frac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \]

Calculating the numerator:

\[ (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6} \]
\[ (\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \]

Adding these gives:

\[ (5 - 2\sqrt{6}) + (5 + 2\sqrt{6}) = 10 \]

Now for the denominator:

\[ (\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = 3 - 2 = 1 \]

Thus, we have:

\[ a + b = \frac{10}{1} = 10 \]

### Step 2: Calculate \(ab\)
Next, we calculate \(ab\):

\[ ab = \left(\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}\right) \left(\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}\right) = \frac{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \]

This simplifies to:

\[ ab = \frac{3 - 2}{1} = 1 \]

### Step 3: Calculate \(\frac{a^2}{b} + \frac{b^2}{a}\)
Using the identity:

\[ \frac{a^2}{b} + \frac{b^2}{a} = \frac{a^3 + b^3}{ab} \]

We know:

\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]

We can express \(a^2 + b^2\) as:

\[ a^2 + b^2 = (a + b)^2 - 2ab = 10^2 - 2 \cdot 1 = 100 - 2 = 98 \]

Thus:

\[ a^3 + b^3 = (a + b)((a^2 + b^2) - ab) = 10(98 - 1) = 10 \cdot 97 = 970 \]

Finally, substituting back into the equation:

\[ \frac{a^2}{b} + \frac{b^2}{a} = \frac{970}{1} = 970 \]

### Final Answer
The value of \(\frac{a^2}{b} + \frac{b^2}{a}\) is:

\[ \boxed{970} \]