Question

If $x^2 + \frac{1}{x^2} = 51$, then what is the value of $x^3 - \frac{1}{x^3}$

• A. 364
• B. 365
• C. 756
• D. 367

Answer 1

Answer: (A)

364

Answer 2

Step 1: Let $t = x + \frac{1}{x}$.
Then

$$t^2 = x^2 + 2 + \frac{1}{x^2} \;\implies\; x^2 + \frac{1}{x^2} = t^2 - 2.$$

Given $x^2 + \tfrac{1}{x^2} = 51$, so

$$t^2 - 2 = 51 \;\implies\; t^2 = 53 \;\implies\; t = \sqrt{53}.$$

Step 2: Compute $x - \tfrac{1}{x}$.

$$\bigl(x - \tfrac{1}{x}\bigr)^2 = x^2 + \tfrac{1}{x^2} - 2 = 51 - 2 = 49 \;\implies\; x - \tfrac{1}{x} = 7.$$

Step 3: Use the identity

$$x^3 - \tfrac{1}{x^3} = \bigl(x - \tfrac{1}{x}\bigr)\,\bigl(x^2 + 1 + \tfrac{1}{x^2}\bigr) = 7 \times (51 + 1) = 7 \times 52 = 364.$$

Thus, the correct answer is 364.