Question

If q is the mean proportional between p and r prove that ${\displaystyle \frac{p^3+q^3+r^3}{p^2{q}^2{r}^2}=\frac{1}{p^3}+\frac{1}{q^3}=\frac{1}{r^3}}$

Answer 1

Since q is the mean proportional between p and r

${\displaystyle q^2=pr}$

L.H.S = ${\displaystyle \frac{p^3+q^3+r^3}{p^2{q}^2{r}^2}}$

${\displaystyle =\frac{p^3+\left(pr\right)q+r^3}{p^3{r}^3}}$

${\displaystyle =\frac{1}{r^3}+\frac{q}{p^2{r}^2}+\frac{1}{p^3}}$

${\displaystyle =\frac{1}{r^3}+\frac{q}{{\left(q^2\right)}^2}+\frac{1}{p^3}}$

${\displaystyle =\frac{1}{r^3}+\frac{1}{q^3}+\frac{1}{p^3}}$

= R.H.S