Solve the following equation by factorization:

$\dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x}$

Given,

$\Rightarrow \dfrac{1}{x - 2} + \dfrac{2}{x - 1} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{(x - 1) + 2(x - 2)}{(x - 2)(x - 1)} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x - 1 + 2x - 4}{x^2 - x - 2x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x - 1 + (2x - 4)}{x^2 - 3x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{3x - 5}{x^2 - 3x + 2} = \dfrac{6}{x} \\[1em] \Rightarrow x(3x - 5) = 6(x^2 - 3x + 2) \\[1em] \Rightarrow 3x^2 - 5x = 6x^2 - 18x + 12 \\[1em] \Rightarrow 6x^2 - 3x^2 - 18x + 5x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 13x + 12 = 0 \\[1em] \Rightarrow 3x^2 - 9x - 4x + 12 = 0 \\[1em] \Rightarrow 3x(x - 3) - 4(x - 3) = 0 \\[1em] \Rightarrow (3x - 4)(x - 3) = 0 \\[1em] \Rightarrow (3x - 4) \text{ or } (x - 3) = 0 \text{ [Using Zero-product rule] } \\[1em] \Rightarrow 3x = 4 \text{ or } x = 3 \\[1em] \Rightarrow x = \dfrac{4}{3} \text{ or } x = 3.$

Hence, $x = \Big{3, \dfrac{4}{3}\Big}$.