# If a$\xi\;b$ are two positive integers,x is their AM and $y\;\xi\;z$ are two GMs then calculate the value of $\frac{y^3+z^3}{xyz}$

$(a)\;\frac{ab^3+a^3b}{abx}\qquad(b)\;2\qquad(c)\;\frac{a^3+b^3}{3ab}\qquad(d)\;\frac{x^3+y^3}{xyz}$

Explanation :
$x=\frac{a+b}{2}\quad\;a,y,z,b\;are\;in\;GP\quad\;b=ar^3\;\quad\;r=(b/a)^\frac{1}{3}$
$y=a(b/a)^\frac{1}{3}\quad\;z=a(b/a)^\frac{2}{3}$
$\frac{y^3+z^3}{xyz}=\frac{[a(b/a)^\frac{1}{3}]^3+[a(b/a)^\frac{2}{3}]^3}{(\frac{a+b}{2})\;[a(b/a)^\frac{1}{3}][a(b/a)^\frac{2}{3}]}$
$=\frac{a^3*b/a+a^3*b^2/a^2}{(\frac{a+b}{2})* [a^2*b/a]}$
$=\frac{a^2b+ab^2}{ab\;(\frac{a+b}{2})}$
$=\frac{ab(a+b)*2}{ab(a+b)}=2.$