Using properties of proportion solve:

$\dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{112(1 - x)}{104(1 + x)}$

Given,

$\Rightarrow \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{112(1 - x)}{104(1 + x)} \\[1em] \Rightarrow \dfrac{x^2 - x + 1}{x^2 + x + 1} = \dfrac{14(1 - x)}{13(1 + x)}.$

Applying componendo and dividendo we get,

$\Rightarrow \dfrac{(x^2 - x + 1) + (x^2 + x + 1)}{(x^2 - x + 1) - (x^2 + x + 1)} = \dfrac{14(1 - x) + 13(1 + x)}{14(1 - x) - 13(1 + x)} \\[1em] \Rightarrow \dfrac{x^2 - x + 1 + x^2 + x + 1}{x^2 - x + 1 - x^2 - x - 1} = \dfrac{14 − 14x + 13 + 13x}{14 − 14x − 13 − 13x} \\[1em] \Rightarrow \dfrac{2x^2 + 2}{-2x} = \dfrac{27 − x}{1 − 27x} \\[1em] \Rightarrow \dfrac{2(x^2 + 1)}{-2x} = \dfrac{27 - x}{1 - 27x} \\[1em] \Rightarrow \dfrac{x^2 + 1}{-x} = \dfrac{27 - x}{1 - 27x}$

⇒ (x² + 1)(1 − 27x) = −x(27 − x)

⇒ x² + 1 − 27x³ − 27x = −27x + x²

⇒ −27x³ + 1 = 0

⇒ 27x³ = 1

⇒ x³ = $\dfrac{1}{27}$

⇒ x = $\sqrt[3]{\dfrac{1}{27}} = \dfrac{1}{3}$

Hence, x = $\dfrac{1}{3}$.