Question

If b is geometric mean of a and c then show that a²b²c² $\left[\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right]=a^3{b}^3{c}^3.$

Answer 1

$$\begin{gathered} \mathrm{Disclaimer}:\mathrm{RHS}\mathrm{should}be{a}^3+b^3+c^3\mathrm{instead}\mathrm{of}{a}^3{b}^3{c}^3 \\ \mathrm{Given}: \\ b\mathrm{is}\mathrm{the}\mathrm{geometric}\mathrm{mean}. \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \frac{a}{b}=\frac{b}{c} \\ \Rightarrow b^2=ac \\ \mathrm{Now}, \\ \mathrm{LHS}=a^2{b}^2{c}^2\left[\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right] \\ =a^2.ac.c^2\left[\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right] \\ =a^3{c}^3\left[\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right] \\ =c^3+\frac{a^3{c}^3}{b^3}+a^3 \\ =c^3+\frac{b^6}{b^3}+a^3 \\ =c^3+b^3+a^3 \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved} \end{gathered}$$