$\text{Given } x + \frac{1}{x} = 2, \text{ we need to find } x^3 + \frac{1}{x^3}.$

Question:

Given: $ x + \frac{1}{x} = 2 $, we need to find $ x^3 + \frac{1}{x^3} $.

Choices:

• 2

• 8

• 14

• 64

Solution and Explanation:

To solve for $ x^3 + \frac{1}{x^3} $, we begin with the given equation:

$ x + \frac{1}{x} = 2 $

We will first square both sides of the equation to find an expression involving $ x^2 + \frac{1}{x^2} $:

$ \left( x + \frac{1}{x} \right)^2 = 2^2 $

Expanding the left side of the equation and simplifying:

$ x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 4 $

Simplifying further, we get:

$ x^2 + 2 + \frac{1}{x^2} = 4 $

Subtracting 2 from both sides, we find:

$ x^2 + \frac{1}{x^2} = 2 $

Next, we need to find an expression for $ x^3 + \frac{1}{x^3} $. For this purpose, we will cube the original equation:

$ \left( x + \frac{1}{x} \right)^3 = 2^3 $

Expanding the left side using the binomial theorem, we get:

$ x^3 + 3 \cdot x \cdot \frac{1}{x} \cdot \left( x + \frac{1}{x} \right) + \frac{1}{x^3} = 8 $

Simplifying the middle term, we have:

$ x^3 + 3 \left( x + \frac{1}{x} \right) + \frac{1}{x^3} = 8 $

Substituting $ x + \frac{1}{x} = 2 $ into the equation:

$ x^3 + \frac{1}{x^3} + 3 \cdot 2 = 8 $

Which simplifies to:

$ x^3 + \frac{1}{x^3} + 6 = 8 $

Solving for $ x^3 + \frac{1}{x^3} $, we subtract 6 from both sides:

$ x^3 + \frac{1}{x^3} = 2 $

Conclusion:

Thus, the correct answer is 2.