Solve the following equation by factorisation:

$\dfrac{1}{x + 6} + \dfrac{1}{x - 10} = \dfrac{3}{x - 4}$

Given,

$\dfrac{1}{x + 6} + \dfrac{1}{x - 10} = \dfrac{3}{x - 4} \\[1em] \Rightarrow \dfrac{x - 10 + x + 6}{(x + 6)(x - 10)} = \dfrac{3}{x - 4} \\[1em] \Rightarrow \dfrac{2x - 4}{(x + 6)(x - 10)} = \dfrac{3}{x - 4} \\[1em] \Rightarrow (2x - 4)(x - 4) = 3(x + 6)(x - 10) \\[1em] \Rightarrow 2x^2 - 8x - 4x + 16 = 3(x^2 - 10x + 6x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3(x^2 - 4x - 60) \\[1em] \Rightarrow 2x^2 - 12x + 16 = 3x^2 - 12x - 180 \\[1em] \Rightarrow 2x^2 - 3x^2 -12x + 12x + 16 + 180 = 0 \\[1em] \Rightarrow -x^2 + 196 = 0 \\[1em] \Rightarrow x^2 = 196 \\[1em] \Rightarrow x = \sqrt{196} \\[1em] x = 14 , -14 \\[1em]$

Hence, the roots of given equation are 14 , -14.