Using properties of proportional, solve the following equation for $^{'} x^{'} : \frac{x^{3} + 3 x}{3 x^{2} + 1} = \frac{341}{91}$

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Solution

$\frac{x^{2} + 3 x}{{3 x}^{2} + 1} = \frac{341}{91}$

using dividendo and componendo rule

$\frac{x^{2} + 3 x + {3 x}^{2} + 1}{x^{2} + 3 x - {3 x}^{2} - 1} = \frac{341 + 91}{341 - 91}$

$\frac{{(x + 1)}^{3}}{{(x - 1)}^{3}} = \frac{432}{250} = (\frac{6}{5})^{3}$

$\frac{(x + 1)}{(x - 1)} = \frac{6}{5}$

using cross-product, we get

$5 (x + 1) = 6 (x - 1)$

$5 x + 5 = 6 x - 6$

$x = 11$ (Answer)

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