If `x-1/x=7` , then `x^3-(1)/(x^3)` is equal to

Options

A. 480
B. 500
C. 364
D. 376

Answer

To solve the problem where \( x - \frac{1}{x} = 7 \) and we need to find \( x^3 - \frac{1}{x^3} \), we can follow these steps:

### Step 1: Start with the given equation
We have:
\[ x - \frac{1}{x} = 7 \]

### Step 2: Cube both sides
To find \( x^3 - \frac{1}{x^3} \), we can use the identity:
\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
Here, let \( a = x \) and \( b = \frac{1}{x} \). Thus:
\[ x^3 - \left(\frac{1}{x}\right)^3 = \left(x - \frac{1}{x}\right) \left(x^2 + x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2\right) \]

### Step 3: Substitute the known values
We know \( x - \frac{1}{x} = 7 \). Now we need to calculate \( x^2 + x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 \):
\[ x^2 + 1 + \frac{1}{x^2} \]

### Step 4: Find \( x^2 + \frac{1}{x^2} \)
We can find \( x^2 + \frac{1}{x^2} \) using the square of \( x - \frac{1}{x} \):
\[ \left(x - \frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} \]
Substituting \( 7 \) into the equation:
\[ 7^2 = x^2 - 2 + \frac{1}{x^2} \]
\[ 49 = x^2 - 2 + \frac{1}{x^2} \]
\[ x^2 + \frac{1}{x^2} = 49 + 2 = 51 \]

### Step 5: Substitute back to find \( x^3 - \frac{1}{x^3} \)
Now substituting back into our earlier expression:
\[ x^3 - \frac{1}{x^3} = \left(x - \frac{1}{x}\right) \left(x^2 + 1 + \frac{1}{x^2}\right) \]
This becomes:
\[ x^3 - \frac{1}{x^3} = 7 \left(51 + 1\right) = 7 \cdot 52 = 364 \]

### Final Answer
Thus, the value of \( x^3 - \frac{1}{x^3} \) is:
\[ \boxed{364} \]