If `(x ^(2))/( y ^(2)) + (y ^(2))/( x ^(2)) = 7,` then the vlaue of `(x ^(3))/( y ^(3)) + (y ^(3))/(x ^(3))` is

Options

A. 16
B. 17
C. 15
D. 18

Answer

To solve the problem, we start with the given equation:

\[ \frac{x^2}{y^2} + \frac{y^2}{x^2} = 7 \]

### Step 1: Let \( a = \frac{x}{y} \)

This means:

\[ \frac{x^2}{y^2} = a^2 \quad \text{and} \quad \frac{y^2}{x^2} = \frac{1}{a^2} \]

Substituting these into the equation gives:

\[ a^2 + \frac{1}{a^2} = 7 \]

### Step 2: Multiply through by \( a^2 \)

This eliminates the fraction:

\[ a^4 - 7a^2 + 1 = 0 \]

### Step 3: Let \( b = a^2 \)

Thus, we can rewrite the equation as:

\[ b^2 - 7b + 1 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( b = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):

Here, \( A = 1, B = -7, C = 1 \):

\[ b = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]
\[ b = \frac{7 \pm \sqrt{49 - 4}}{2} \]
\[ b = \frac{7 \pm \sqrt{45}}{2} \]
\[ b = \frac{7 \pm 3\sqrt{5}}{2} \]

### Step 5: Find \( a^3 + \frac{1}{a^3} \)

We know that:

\[ a^3 + \frac{1}{a^3} = \left(a + \frac{1}{a}\right)\left(a^2 - 1 + \frac{1}{a^2}\right) \]

We already have \( a^2 + \frac{1}{a^2} = 7 \). Now we need \( a + \frac{1}{a} \).

### Step 6: Find \( a + \frac{1}{a} \)

Using the identity:

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

We can express this as:

\[ \left(a + \frac{1}{a}\right)^2 = 7 + 2 = 9 \]

Thus:

\[ a + \frac{1}{a} = 3 \]

### Step 7: Calculate \( a^3 + \frac{1}{a^3} \)

Now we can substitute back:

\[ a^3 + \frac{1}{a^3} = (a + \frac{1}{a})\left(a^2 + \frac{1}{a^2} - 1\right) \]
\[ = 3(7 - 1) = 3 \cdot 6 = 18 \]

### Final Answer

Thus, the value of \( \frac{x^3}{y^3} + \frac{y^3}{x^3} \) is:

\[ \boxed{18} \]