Question

If \( x + \frac{1}{x} = 2 \), then \( x^3 + \frac{1}{x^3} = \)

• A. 64
• B. 8
• C. 2
• D. 1

Hint:

A useful trick: If \(x + \frac{1}{x} = 2\), then \(x\) must be 1. You can check this: \(1 + \frac{1}{1} = 1 + 1 = 2\). Once you know \(x=1\), then \(x^n + \frac{1}{x^n}\) will always be \(1^n + \frac{1}{1^n} = 1+1 = 2\) for any integer \(n\). So, \(x^3 + \frac{1}{x^3} = 2\).

Answer 1

The Correct Option is C

Concept: This problem can be solved in two ways: by finding the value of \(x\) first, or by using an algebraic identity.

Method 1: Finding the value of \(x\) Given: \( x + \frac{1}{x} = 2 \) Multiply the entire equation by \(x\) (assuming \(x \neq 0\)): \[ x(x) + x\left(\frac{1}{x}\right) = 2(x) \] \[ x^2 + 1 = 2x \] Rearrange into a quadratic equation: \[ x^2 - 2x + 1 = 0 \] This is a perfect square trinomial: \((x-1)^2 = 0\). Taking the square root of both sides: \[ x - 1 = 0 \] \[ x = 1 \] Now that we have \(x=1\), we can find the value of \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = (1)^3 + \frac{1}{(1)^3} = 1 + \frac{1}{1} = 1 + 1 = 2 \] Method 2: Using the algebraic identity for \((a+b)^3\) We know the identity: \((a+b)^3 = a^3 + b^3 + 3ab(a+b)\). Let \(a = x\) and \(b = \frac{1}{x}\). Then \(a+b = x + \frac{1}{x}\). And \(a^3 + b^3 = x^3 + \left(\frac{1}{x}\right)^3 = x^3 + \frac{1}{x^3}\). Also, \(ab = x \cdot \frac{1}{x} = 1\). Substitute these into the identity: \[ \left(x + \frac{1}{x}\right)^3 = \left(x^3 + \frac{1}{x^3}\right) + 3\left(x \cdot \frac{1}{x}\right)\left(x + \frac{1}{x}\right) \] \[ \left(x + \frac{1}{x}\right)^3 = \left(x^3 + \frac{1}{x^3}\right) + 3(1)\left(x + \frac{1}{x}\right) \] We are given \( x + \frac{1}{x} = 2 \). Substitute this value: \[ (2)^3 = \left(x^3 + \frac{1}{x^3}\right) + 3(2) \] \[ 8 = \left(x^3 + \frac{1}{x^3}\right) + 6 \] Now, solve for \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = 8 - 6 \] \[ x^3 + \frac{1}{x^3} = 2 \] Both methods yield the same result.