If the mean of x and 1/x is M, then the mean of x³ and 1/x³ is :

Given $\frac{x + 1/x}{2} = M \implies x + \frac{1}{x} = 2M$. Using the algebraic identity $x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)^3 - 3\left(x + \frac{1}{x}\right) = (2M)^3 - 3(2M) = 8M^3 - 6M$, the arithmetic mean of $x^3$ and $\frac{1}{x^3}$ is $\frac{8M^3 - 6M}{2} = 4M^3 - 3M = M(4M^2 - 3)$.