Given the equation \( x - \frac{1}{x} = 4 \), we are asked to find \( x^3 + \frac{1}{x^3} \).

### Step 1: Square both sides of the equation
First, square both sides of the equation \( x - \frac{1}{x} = 4 \):

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

Expanding the left side:

\[
x^2 - 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 16
\]

Simplifying:

\[
x^2 - 2 + \frac{1}{x^2} = 16
\]

Now, isolate \( x^2 + \frac{1}{x^2} \):

\[
x^2 + \frac{1}{x^2} = 16 + 2 = 18
\]

### Step 2: Use the identity for cubes
To find \( x^3 + \frac{1}{x^3} \), use the identity:

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

We already know:

- \( x^2 + \frac{1}{x^2} = 18 \)
- \( x - \frac{1}{x} = 4 \)

Now, find \( x + \frac{1}{x} \):

\[
x + \frac{1}{x} = \sqrt{ \left( x - \frac{1}{x} \right)^2 + 4 } = \sqrt{16 + 4} = \sqrt{20}
\]

So \( x + \frac{1}{x} \approx 4.47 \). Plugging this back:

\[
x^3 + \frac{1}{x^3} = (

Learn how to solve for x³ + 1/x³ given x - 1/x = 4. This step-by-step solution demonstrates algebraic techniques, including binomial expansion and exponent manipulation, suitable for Grades 9-10 students.

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