If (3x² + 2y²) : (3x² - 2y²) = 11 : 9, find the value of $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4}.$

Given,

$\dfrac{3x^2 + 2y^2}{3x^2 - 2y^2} = \dfrac{11}{9}$

Applying componendo and dividendo to above equation,

$\Rightarrow \dfrac{3x^2 + 2y^2 + 3x^2 - 2y^2}{3x^2 + 2y^2 - 3x^2 + 2y^2} = \dfrac{11 + 9}{11 - 9} \\[1em] \Rightarrow \dfrac{6x^2}{4y^2} = \dfrac{20}{2} \\[1em] \Rightarrow \dfrac{3x^2}{2y^2} = 10 \\[1em] \Rightarrow \dfrac{x^2}{y^2} = \dfrac{20}{3}$

Putting value of $\dfrac{x^2}{y^2}$ = $\dfrac{20}{3}$ in $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4}$,

$\Rightarrow \dfrac{3\big(\dfrac{x^2}{y^2}\big)^2 + 25}{3\big(\dfrac{x^2}{y^2}\big)^2 - 25} \\[1em] \Rightarrow \dfrac{3\big(\dfrac{400}{9}\big) + 25}{3\big(\dfrac{400}{9}\big) - 25} \\[1em] \Rightarrow \dfrac{\dfrac{400}{3} + 25}{\dfrac{400}{3} - 25} \\[1em] \Rightarrow \dfrac{\dfrac{400 + 75}{3}}{\dfrac{400 - 75}{3}} \\[1em] \Rightarrow \dfrac{475}{325} \\[1em] \Rightarrow \dfrac{19}{13}.$

Hence, the value of $\dfrac{3x^4 + 25y^4}{3x^4 - 25y^4} \text{ is } \dfrac{19}{13}.$