ĐK: $x \neq \pm 1$
Khi đó, ta có
$\dfrac{(x+1)(x-1)}{(x-1)(x^2 + x + 1)} - \dfrac{(x-1)(x+1)}{(x+1)(x^2 - x +1)} = \dfrac{2(x+2)^2}{(x^3-1)(x^3+1)}$
$<-> \dfrac{x^2 - 1}{x^3-1} - \dfrac{x^2-1}{x^3 + 1} = \dfrac{2(x+2)^2}{(x^3-1)(x^3+1)}$
$<-> (x^2-1)(x^3+1) - (x^2-1)(x^3-1) = 2(x+2)^2$
$<-> (x^2-1)(x^3+1 - x^3 + 1) = 2(x+2)^2$
$<-> x^2-1 = x^2 + 4x + 4$
$<-> 4x = -5$
$<-> x = -\dfrac{5}{4}$
