Find the compounded ratio of:

(i) 2 : 3 and 4 : 9

(ii) 4 : 5, 5 : 7 and 9 : 11

(iii) (a - b) : (a + b), (a + b)² : (a² + b²) and (a⁴ - b⁴) : (a² - b²)²

(i) The compounded ratio of 2 : 3 and 4 : 9 is,

$= \dfrac{2}{3} \times \dfrac{4}{9} \\[0.5em] = \dfrac{8}{27}$

Hence, the compounded ratio is 8 : 27.

(ii) The compounded ratio of 4 : 5, 5 : 7 and 9 : 11 is,

$= \dfrac{4}{5} \times \dfrac{5}{7} \times \dfrac{9}{11} \\[0.5em] = \dfrac{180}{385} \\[0.5em]$

Dividing numerator and denominator by 5, we get:

$\dfrac{\overset{36}{\bcancel{180}}}{\underset{77}{\bcancel{385}}} = \dfrac{36}{77}$

Hence, the compounded ratio is 36 : 77.

(iii) The compounded ratio of (a - b) : (a + b), (a + b)² : (a² + b²) and (a⁴ - b⁴) : (a² - b²)² is,

$= \dfrac{(a - b)}{(a + b)} \times \dfrac{(a + b)^2}{(a^2 + b^2)} \times \dfrac{(a^4 - b^4)}{(a^2 - b^2)^2} \\[0.5em] = \dfrac{(a - b)}{\bcancel{(a + b)}} \times \dfrac{\bcancel{(a + b)}(a + b)}{\bcancel{(a^2 + b^2)}} \times \dfrac{\bcancel{(a^2 - b^2)}\bcancel{(a^2 + b^2)}}{\bcancel{(a^2 - b^2)}(a^2 - b^2)} \\[0.5em] = \dfrac{(a - b)(a + b)}{(a^2 - b^2)} \\[0.5em] = \dfrac{(a^2 - b^2)}{(a^2 - b^2)} \\[0.5em] = \dfrac{1}{1}$

Hence, the compounded ratio is 1 : 1.