Question

The mean of \( x \) and \( \frac{1}{x} \) is \( M \). Find the mean of \( x^3 \) and \( \frac{1}{x^3} \).

Hint:

To find the mean of expressions involving powers of \( x \) and \( \frac{1}{x} \), use algebraic identities to simplify the problem.

Answer 1

The mean of two numbers is the sum of the numbers divided by 2. Given that the mean of \( x \) and \( \frac{1}{x} \) is \( M \), we can write: \[ M = \frac{x + \frac{1}{x}}{2}. \] We need to find the mean of \( x^3 \) and \( \frac{1}{x^3} \). The mean is: \[ \text{Mean} = \frac{x^3 + \frac{1}{x^3}}{2}. \] Step 1: Use the identity for \( x + \frac{1}{x} \). We square the equation for the mean of \( x \) and \( \frac{1}{x} \): \[ \left( x + \frac{1}{x} \right)^2 = x^2 + 2 + \frac{1}{x^2}. \] Thus, we have: \[ M^2 = x^2 + 2 + \frac{1}{x^2}. \] Step 2: Find \( x^3 + \frac{1}{x^3} \). Now, we use the identity for \( x^3 + \frac{1}{x^3} \): \[ \left( x + \frac{1}{x} \right)^3 = x^3 + 3x + \frac{3}{x} + \frac{1}{x^3}. \] Thus: \[ \left( x + \frac{1}{x} \right)^3 = x^3 + \frac{1}{x^3} + 3\left( x + \frac{1}{x} \right). \] Substitute \( x + \frac{1}{x} = 2M \): \[ (2M)^3 = x^3 + \frac{1}{x^3} + 3(2M), \] \[ 8M^3 = x^3 + \frac{1}{x^3} + 6M. \] Solving for \( x^3 + \frac{1}{x^3} \): \[ x^3 + \frac{1}{x^3} = 8M^3 - 6M. \] Step 3: Find the mean of \( x^3 \) and \( \frac{1}{x^3} \). Finally, the mean of \( x^3 \) and \( \frac{1}{x^3} \) is: \[ \frac{x^3 + \frac{1}{x^3}}{2} = \frac{8M^3 - 6M}{2} = 4M^3 - 3M. \]
Conclusion:
The mean of \( x^3 \) and \( \frac{1}{x^3} \) is \( 4M^3 - 3M \).