If ` x = sqrt(3) - (1)/( sqrt(3)) and y = sqrt(3) + (1)/( sqrt(3))` then the value of ` (x^(2))/(y) + (y^(2))/(x)` is

Options

A. `sqrt(3)`
B. ` 3 sqrt(3)`
C. `16 sqrt(3)`
D. ` 2 sqrt(3)`

Answer

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

### Step 1: Calculate \( x^2 \) and \( y^2 \)

First, we calculate \( x^2 \):
\[ x = \sqrt{3} - \frac{1}{\sqrt{3}} \]
\[ x^2 = \left(\sqrt{3} - \frac{1}{\sqrt{3}}\right)^2 = (\sqrt{3})^2 - 2 \cdot \sqrt{3} \cdot \frac{1}{\sqrt{3}} + \left(\frac{1}{\sqrt{3}}\right)^2 \]
\[ = 3 - 2 + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3} \]

Now calculate \( y^2 \):
\[ y = \sqrt{3} + \frac{1}{\sqrt{3}} \]
\[ y^2 = \left(\sqrt{3} + \frac{1}{\sqrt{3}}\right)^2 = (\sqrt{3})^2 + 2 \cdot \sqrt{3} \cdot \frac{1}{\sqrt{3}} + \left(\frac{1}{\sqrt{3}}\right)^2 \]
\[ = 3 + 2 + \frac{1}{3} = 5 + \frac{1}{3} = \frac{16}{3} \]

### Step 2: Substitute \( x^2 \) and \( y^2 \) into the expression

Now substitute \( x^2 \) and \( y^2 \) into the expression \( \frac{x^2}{y} + \frac{y^2}{x} \):
\[ \frac{x^2}{y} + \frac{y^2}{x} = \frac{\frac{4}{3}}{y} + \frac{\frac{16}{3}}{x} \]

### Step 3: Calculate \( \frac{4}{3y} \) and \( \frac{16}{3x} \)

Now we need to calculate \( y \) and \( x \):
\[ y = \sqrt{3} + \frac{1}{\sqrt{3}} = \frac{3 + 1}{\sqrt{3}} = \frac{4}{\sqrt{3}} \]
\[ x = \sqrt{3} - \frac{1}{\sqrt{3}} = \frac{3 - 1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \]

Now substitute \( y \) and \( x \) into the fractions:
\[ \frac{4}{3y} = \frac{4}{3 \cdot \frac{4}{\sqrt{3}}} = \frac{4 \sqrt{3}}{12} = \frac{\sqrt{3}}{3} \]
\[ \frac{16}{3x} = \frac{16}{3 \cdot \frac{2}{\sqrt{3}}} = \frac{16 \sqrt{3}}{6} = \frac{8 \sqrt{3}}{3} \]

### Step 4: Combine the fractions

Now combine the two results:
\[ \frac{x^2}{y} + \frac{y^2}{x} = \frac{\sqrt{3}}{3} + \frac{8\sqrt{3}}{3} = \frac{9\sqrt{3}}{3} = 3\sqrt{3} \]

### Final Answer
Thus, the value of \( \frac{x^2}{y} + \frac{y^2}{x} \) is \( 3\sqrt{3} \).

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