If `(x^(32)-1)/(x^(16))=5` then find the value of `(x^(96)-1)/(x^(48))`.

Options

A. 40
B. 70
C. 140
D. 180

Answer

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

\[ \frac{x^{32} - 1}{x^{16}} = 5 \]

### Step 1: Rewrite the equation
We can rewrite the left-hand side of the equation:

\[ \frac{x^{32} - 1}{x^{16}} = \frac{x^{32}}{x^{16}} - \frac{1}{x^{16}} = x^{16} - \frac{1}{x^{16}} = 5 \]

### Step 2: Let \( y = x^{16} \)
Now, we let \( y = x^{16} \). Substituting this into our equation gives:

\[ y - \frac{1}{y} = 5 \]

### Step 3: Multiply through by \( y \)
To eliminate the fraction, we multiply both sides by \( y \):

\[ y^2 - 1 = 5y \]

### Step 4: Rearrange the equation
Rearranging gives us a quadratic equation:

\[ y^2 - 5y - 1 = 0 \]

### Step 5: Solve the quadratic equation
We can solve for \( y \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 1, b = -5, c = -1 \):

\[ y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2} \]

### Step 6: Find \( \frac{x^{96} - 1}{x^{48}} \)
Next, we need to find the value of:

\[ \frac{x^{96} - 1}{x^{48}} \]

Using the substitution \( y = x^{16} \):

\[ x^{96} = (x^{16})^6 = y^6 \quad \text{and} \quad x^{48} = (x^{16})^3 = y^3 \]

Thus, we have:

\[ \frac{x^{96} - 1}{x^{48}} = \frac{y^6 - 1}{y^3} \]

### Step 7: Simplify the expression
This can be simplified as:

\[ \frac{y^6 - 1}{y^3} = y^3 - \frac{1}{y^3} \]

### Step 8: Find \( y^3 - \frac{1}{y^3} \)
To find \( y^3 - \frac{1}{y^3} \), we can use the identity:

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

We already know \( y - \frac{1}{y} = 5 \). Now we need to find \( y^2 + 1 + \frac{1}{y^2} \).

### Step 9: Calculate \( y^2 + \frac{1}{y^2} \)
Using the square of \( y - \frac{1}{y} \):

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

This gives:

\[ 5^2 = y^2 - 2 + \frac{1}{y^2} \implies 25 = y^2 - 2 + \frac{1}{y^2} \implies y^2 + \frac{1}{y^2} = 27 \]

### Step 10: Substitute back to find \( y^3 - \frac{1}{y^3} \)
Now we substitute back:

\[ y^3 - \frac{1}{y^3} = 5(27 + 1) = 5 \times 28 = 140 \]

### Final Answer
Thus, the value of \( \frac{x^{96} - 1}{x^{48}} \) is:

\[ \boxed{140} \]