If $\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y}$ (x ≠ 0, y ≠ 0, x ≠ y), then find the value of x³ - y³

Given,

$\dfrac{1}{x + y} = \dfrac{1}{x} + \dfrac{1}{y}$

$\Rightarrow \dfrac{1}{x + y} = \dfrac{y + x}{xy} \\[1em] \Rightarrow xy = (x + y)(x + y) \\[1em] \Rightarrow xy = x^2 + y^2 + 2xy \\[1em] \Rightarrow 0 = x^2 + y^2 + 2xy - xy \\[1em] \Rightarrow x^2 + xy + y^2 = 0$

Using identity,

x³ - y³ = (x - y)(x² + xy +y²)

x³ - y³ = (x - y)(0)

x³ - y³ = 0

Hence, x³ - y³ = 0.