We are tasked with evaluating the expression \(\frac{x^6 - 1}{x^3}\), given that \(x - \frac{1}{x} = 4\).

### Step 1: Express \(x^2 + \frac{1}{x^2}\)
We can square both sides of the given equation \(x - \frac{1}{x} = 4\):

\[
\left( x - \frac{1}{x} \right)^2 = 4^2
\]
\[
x^2 - 2 + \frac{1}{x^2} = 16
\]
\[
x^2 + \frac{1}{x^2} = 18
\]

### Step 2: Express \(x^3 + \frac{1}{x^3}\)
Next, we want to cube both sides of the original equation \(x - \frac{1}{x} = 4\):

\[
\left( x - \frac{1}{x} \right)^3 = 4^3
\]
Using the identity \(\left( x - \frac{1}{x} \right)^3 = x^3 - 3x + \frac{1}{x^3} + \frac{3}{x}\), we can rewrite this as:
\[
x^3 - 3\left(x - \frac{1}{x}\right) + \frac{1}{x^3} = 64
\]
Substitute \(x - \frac{1}{x} = 4\):
\[
x^3 - 12 + \frac{1}{x^3} = 64
\]
\[
x^3 + \frac{1}{x^3} = 76
\]

### Step 3: Simplify \(\frac{x^6 - 1}{x^3}\)
Now, we express \(\frac{x^6 - 1}{x^3}\). Notice that:
\[
\frac{x^6 - 1}{x^3} = x^3 - \frac{1}{x^3}
\]
Since we know that \(x^3 + \frac{1}{x^3} = 76\), we can subtract to find \(x^3 - \frac{1}{x^3}\):
\[
x^3 - \frac{1}{x^3} = \sqrt{(x^3 + \frac{1}{x^3})^2 - 4}
\]




Learn how to evaluate the expression (x^6 - 1)/x^3 when x - 1/x = 4. This solution includes key algebraic manipulations, cubing, and squaring techniques, suitable for high school students in Grades 10-12.

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