Question

Let 'x' be the arithmetic mean and y,z be the geometric means between any two geometric numbers.The value of $\frac{(y^3+z^3)}{\left(xyz\right)}$ is?

Answer 1

Let a and b be two positive numbers. Then
x = A.M. of a and b $=\frac{a+b}{2}$
and a, y, z, b will be in G.P. with
common ratio $\left(r\right)={\left(\frac{b}{a}\right)}^{\frac{1}{2+1}}={\left(\frac{b}{a}\right)}^{\frac{1}{3}}$
$\Rightarrow y=ar=a{\left(\frac{b}{a}\right)}^{\frac{1}{3}}=a^{\frac{2}{3}}{b}^{\frac{1}{3}}$
$andz=a{r}^2=a^{\frac{1}{3}}{b}^{\frac{2}{3}}$
$\therefore y^3+z^3=a^{2}b+a{b}^2=ab(a+b)$
= 2abx
and yz = ab
$\Rightarrow y^3+z^3=2xyz$
$\Rightarrow \frac{y^3+z^3}{xyz}=2$