If b is the mean proportion between a and c, show that:

$\dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}}$

Given,

Since b is the mean proportional between a and c, we have

⇒ a : b :: b : c

⇒ $\dfrac{a}{b} = \dfrac{b}{c}$

⇒ b² = ac

Substituting value of b² in $\dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}}$, we get :

$\Rightarrow \dfrac{a^{4} + a^{2}b^{2} + (b^{2})^2}{(b^{2})^2 + b^{2}c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^{4} + a^{2}.ac + (ac)^2}{(ac)^2 + (ac).c^{2} + c^{4}} \\[1em] \Rightarrow \dfrac{a^2(a^{2} + ac + c^2)}{c^2(a^2 + ac + c^2)} \\[1em] \Rightarrow \dfrac{a^2}{c^2}.$

Hence, proved that $\dfrac{a^{4} + a^{2}b^{2} + b^{4}}{b^{4} + b^{2}c^{2} + c^{4}} = \dfrac{a^{2}}{c^{2}}$.