Given that \( x + \frac{1}{x} = 5 \), we are tasked with finding the value of \( x^5 - \frac{1}{x^5} \).

### Step 1: Square both sides
To proceed, first square both sides of the given equation \( x + \frac{1}{x} = 5 \):

\[
\left( x + \frac{1}{x} \right)^2 = 5^2
\]

Expanding the left side:

\[
x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 25
\]

Since \( 2 \cdot x \cdot \frac{1}{x} = 2 \), we simplify:

\[
x^2 + 2 + \frac{1}{x^2} = 25
\]

Subtract 2 from both sides:

\[
x^2 + \frac{1}{x^2} = 23
\]

### Step 2: Cube both sides of \( x + \frac{1}{x} = 5 \)
Now, cube both sides of the equation \( x + \frac{1}{x} = 5 \):

\[
\left( x + \frac{1}{x} \right)^3 = 5^3
\]

Expanding the left side using the identity \( (a+b)^3 = a^3 + b^3 + 3ab(a+b) \):

\[
x^3 + \frac{1}{x^3} + 3 \left( x + \frac{1}{x} \right) = 125
\]

Since \( x + \frac{1}{x} = 5 \), substitute it into the equation:

\[
x^3 + \frac{1}{x^3} + 3 \cdot 5 = 125
\]

Simplifying:

\[
x^3 + \frac{1}{x^3} + 15 = 125
\]

Subtract 15 from both sides:

\[
x^3 + \frac{1}{x^3} = 110
\]

### Step 3: Use the identity for \( x^5 + \frac{1}{x^5} \)
To find \( x^5 - \frac{1}{x^5} \), use the identity:

\[
\left( x + \frac{1}{x} \right)\left( x^4 - x^3 + x^2 - x + 1 \right) = x^5 - \frac{1}{x^5}
\]


If x + 1/x = 5, then find the value of x^5 - 1/x^5.

Discover how to solve the equation x + 1/x = 5 and find the value of x^5 - 1/x^5. This problem involves algebraic identities and exponent rules, perfect for students in Grades 9-11.

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