Question

If the mean of x and $\frac{1}{x}$ is M, then the mean of x${}^3$ and $\frac{1}{x^3}$ is

• A. $\frac{M(M^2-3)}{2}$
• B. $M(4M^2-3)$
• C. $M^3$
• D. $M^3+3$

Answer 1

The correct option is B $M(4M^2-3)$

Given,

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

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

On cubing both sides, we get

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

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

$8M^3=x^3+\frac{1}{x^3}+3\left(2M\right)$

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

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

$\frac{x^3+\frac{1}{x^3}}{2}=M(4M^2-3)$

Hence, this is the answer.