If m = $\sqrt[3]{15} \text{ and } n = \sqrt[3]{14}$, find the value of m - n - $\dfrac{1}{m^2 + mn + n^2}$.

Given,

$\Rightarrow m = \sqrt[3]{15} \text{ and } n = \sqrt[3]{14} \\[1em] \Rightarrow m = (15)^{\dfrac{1}{3}} \text{ and } n = (14)^{\dfrac{1}{3}}$

Cubing both sides, we get :

$\Rightarrow m^3 = [(15)^{\dfrac{1}{3}}]^3 \text{ and } n^3 = [(14)^{\dfrac{1}{3}}]^3 \\[1em] \Rightarrow m^3 = (15)^{\dfrac{1}{3} \times 3} \text{ and } n^3 = (14)^{\dfrac{1}{3} \times 3} \\[1em] \Rightarrow m^3 = 15 \text{ and } n^3 = 14.$

Simplifying the expression $m - n - \dfrac{1}{m^2 + mn + n^2}$, we get :

$\Rightarrow \dfrac{m(m^2 + mn + n^2) - n(m^2 + mn + n^2) - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 + m^2n + mn^2 - nm^2 - mn^2 - n^3 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{m^3 - n^3 - 1}{m^2 + mn + n^2} \\[1em]$

Substituting value of m³ and n³ in above equation, we get :

$\Rightarrow \dfrac{15 - 14 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{1 - 1}{m^2 + mn + n^2} \\[1em] \Rightarrow \dfrac{0}{m^2 + mn + n^2} \\[1em] \Rightarrow 0.$

Hence, $m - n - \dfrac{1}{m^2 + mn + n^2} = 0$