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

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

Given,

b is the mean proportion between a and c

$\therefore \dfrac{a}{b} = \dfrac{b}{c}$

⇒ b² = ac

Substituting b² = ac in L.H.S. of the equation $\dfrac{a^4 + a^2b^2 + b^4}{b^4 + b^2c^2 + c^4} = \dfrac{a^2}{c^2}$ we get,

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

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