If A be one A.M. and p, q be two G.M.'s between two numbers,

(a) $\frac{p^{3}+q^{3}}{p q}$

Let the two positive numbers be $a$ and $b$.

$a, A$ and $b$ are in A.P.

$\therefore 2 A=a+b$   ...(i)

Also, $a, p, q$ and $b$ are in G.P.

$\therefore r=\left(\frac{b}{a}\right)^{\frac{1}{3}}$

Again, $p=a r$ and $q=a r^{2}$   ...(ii)

Now, $2 A=a+b \quad[$ From (i) $]$

$=a+a\left(\frac{b}{a}\right)$

$=a+a\left(\left(\frac{b}{a}\right)^{\frac{1}{3}}\right)^{3}$

$=a+a r^{3}$

$=\frac{(a r)^{2}}{a r^{2}}+\frac{\left(a r^{2}\right)^{2}}{a r}$

$=\frac{p^{2}}{q}+\frac{q^{2}}{p} \quad[\mathrm{U} \sin g(\mathrm{ii})]$

$=\frac{p^{3}+q^{3}}{p q}$