Given x = $\sqrt[3]{65}$ and y = $\sqrt[3]{64}$, find the value of

$(x - y) - \dfrac{1}{x^2 + xy + y^2}$

Given,

x = $\sqrt[3]{65}$

y = $\sqrt[3]{64}$

By formula,

x³ - y³ = (x - y) (x² + xy + y²)

x³ - y³ = $(\sqrt[3]{65})^3$ - $(\sqrt[3]{64})^3$ = 65 - 64 = 1

∴ 1 = (x - y) (x² + xy + y²)……(1)

Solving,

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

Hence, $(x - y) - \dfrac{1}{x^2 + xy + y^2}$ = 0.