If $\;$ $\dfrac{x^2 + x + 1}{y^2 + y + 1} = \dfrac{x^2 - x + 1}{y^2 - y + 1}$, $\;$ then show that $\;$ $xy = 1$.
Given: $\;$ $\dfrac{x^2 + x + 1}{y^2 + y +1} = \dfrac{x^2 - x + 1}{y^2 - y + 1}$
i.e. $\;$ $\dfrac{x^2 + x + 1}{x^2 - x + 1} = \dfrac{y^2 + y + 1}{y^2 - y + 1}$
By componendo-dividendo we have,
$\dfrac{\left(x^2 + x + 1\right) + \left(x^2 - x + 1\right)}{\left(x^2 + x + 1\right)- \left(x^2 - x + 1\right)} = \dfrac{\left(y^2 + y + 1\right) + \left(y^2 - y + 1\right)}{\left(y^2 + y + 1\right)- \left(y^2 - y + 1\right)}$
i.e. $\;$ $\dfrac{2 x^2 + 2}{2x} = \dfrac{2 y^2 + 2}{2 y}$
i.e. $\;$ $\dfrac{x^2 + 1}{x} = \dfrac{y^2 + 1}{y}$
i.e. $\;$ $x^2 y + y = y^2 x + x$
i.e. $\;$ $x^2 y - y^2 x = x - y$
i.e. $\;$ $xy \left(x - y\right) = x - y$
i.e. $\;$ $xy = 1$ $\;$ provided $\;$ $x - y \neq 0$ $\;$ i.e. $\;$ $x \neq y$
Hence proved.