If `x=sqrta+(1)/(sqrta), y=sqrta-(1)/(sqrta)` then the value of `x^(4)+y^(4)-2x^(2)y^(2)` is

Options

A. 16
B. 20
C. 10
D. 5

Answer

To solve the problem, we start with the given expressions for \( x \) and \( y \):

\[ x = \sqrt{a} + \frac{1}{\sqrt{a}} \]
\[ y = \sqrt{a} - \frac{1}{\sqrt{a}} \]

We need to find the value of \( x^4 + y^4 - 2x^2y^2 \).

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

First, we calculate \( x^2 \) and \( y^2 \):

\[ x^2 = \left( \sqrt{a} + \frac{1}{\sqrt{a}} \right)^2 = a + 2 + \frac{1}{a} \]

\[ y^2 = \left( \sqrt{a} - \frac{1}{\sqrt{a}} \right)^2 = a - 2 + \frac{1}{a} \]

### Step 2: Calculate \( x^2 + y^2 \)

Next, we find \( x^2 + y^2 \):

\[ x^2 + y^2 = \left( a + 2 + \frac{1}{a} \right) + \left( a - 2 + \frac{1}{a} \right) \]
\[ = 2a + 2\cdot\frac{1}{a} = 2\left(a + \frac{1}{a}\right) \]

### Step 3: Calculate \( x^2y^2 \)

Now, we calculate \( x^2y^2 \):

\[ x^2y^2 = \left( a + 2 + \frac{1}{a} \right)\left( a - 2 + \frac{1}{a} \right) \]

Using the difference of squares:

\[ = \left( a + \frac{1}{a} \right)^2 - 2^2 = \left( a + \frac{1}{a} \right)^2 - 4 \]

### Step 4: Calculate \( x^4 + y^4 \)

Now we can use the identity \( x^4 + y^4 = (x^2 + y^2)^2 - 2x^2y^2 \):

\[ x^4 + y^4 = \left( 2\left(a + \frac{1}{a}\right) \right)^2 - 2\left( \left( a + \frac{1}{a} \right)^2 - 4 \right) \]
\[ = 4\left(a + \frac{1}{a}\right)^2 - 2\left( \left( a + \frac{1}{a} \right)^2 - 4 \right) \]
\[ = 4\left(a + \frac{1}{a}\right)^2 - 2\left(a + \frac{1}{a}\right)^2 + 8 \]
\[ = 2\left(a + \frac{1}{a}\right)^2 + 8 \]

### Step 5: Combine to find \( x^4 + y^4 - 2x^2y^2 \)

Finally, we substitute back into the expression we need to evaluate:

\[ x^4 + y^4 - 2x^2y^2 = \left( 2\left(a + \frac{1}{a}\right)^2 + 8 \right) - 2\left( \left( a + \frac{1}{a} \right)^2 - 4 \right) \]
\[ = 2\left(a + \frac{1}{a}\right)^2 + 8 - 2\left(a + \frac{1}{a}\right)^2 + 8 \]
\[ = 16 \]

Thus, the value of \( x^4 + y^4 - 2x^2y^2 \) is:

\[ \boxed{16} \]