Question

if the mean of y and 1/y is m then what is the mean of y^3 and 1/y^3

Answer 1

$\textbf{Given:}$

$\text{Mean of y and \frac{1}{y} is m}$

$\implies\frac{y+\frac{1}{y}}{2}=m$

$\implies\bf\,y+\frac{1}{y}=2m$

$\text{Using}$

$\boxed{\bf\,a^3+b^3=(a+b)^3-3ab(a+b)}$

$y^3+\frac{1}{y^3}=(y+\frac{1}{y})^3-3y(\frac{1}{y})(y+\frac{1}{y})}$

$y^3+\frac{1}{y^3}=(y+\frac{1}{y})^3-3(y+\frac{1}{y})}$

$=(2m)^3-3(2m)$

$=8m^3-6m$

$\textbf{Mean of y^3 and \frac{1}{y^3} }$

$=\frac{y^3+\frac{1}{y^3}}{2}$

$=\frac{8m^3-6m}{2}$

$=4m^3-3m$