asymptote of Lamé's cubic

We will show that the Lamé's cubic

where $a$ is a positive constant, has the line $$y = \underbrace{-x}_{g(x)}$$ as its asymptote.

Because the equation (1) of the curve is symmetric with respect to $x$ and $y$ , the curve is symmetric about the line $y = x$ . From the solved form

of (1) we see that every real value of $x$ gives one point of the curve.

The difference $\Delta = f(x)\!-\!g(x)$ represents the distance of a point $(x,\,y)$ of the curve and the point of the asserted asymptote $y = -x$ with the same abscissa $x$ . We multiply the numerator and denominator with the expression $(\sqrt[3]{a^3-x^3})^2-x\sqrt[3]{a^3-x^3}+x^2$ for being able to utilise the polynomial formula $$(u+v)(u^2-uv+v^2) = u^3+v^3,$$ getting

Thus, $\displaystyle \Delta \to \frac{a^3}{\infty+\infty+\infty} = 0$ when $ |x| \to \infty$ (see the improper limits). According to the definition of asymptote, the line $y = -x$ is asymptote of Lamé's cubic.