I don't think I can do it

This is a Folium of Descartes , $x^3+y^3=3axy$ , with $a=\frac{1729}{3}$ . The loop $C$ of this curve (i.e., the part in the first quadrant) can be parametrized by $x=\frac{3at}{1+t^3}, y=\frac{3at^2}{1+t^3}$ for $t\geq 0$ . Thus the area enclosed by $C$ is $\int_{C}xdy=-9a^2\int_{0}^{\infty}\frac{t^2(t^3-2)}{(1+t^3)^3}dt=-9a^2\left[\frac{1-2t^3}{6(1+t^3)^2}\right]_{0}^{\infty}=\frac{3a^2}{2}=\frac{1729^2}{6}$

The sum we seek is $\boxed{1735}$

We can also solve this problem by converting the equation $x^3+y^3=3axy$ into polar coordinates. So we get $r=\frac{3a \sin\theta \cos \theta}{\cos^3 \theta +\sin^3\theta}.$ Dividing both the numerator and denominator by $\cos^3\theta$ we obtain that $r=\frac{3a \sec\theta \tan \theta}{1 +\tan^3\theta}.$ Now the given area can be expressed as $A=\frac{1}{2}\int_0^{\pi/2}r^2d\theta=\frac{1}{2}\int_0^{\pi/2}\frac{9a^2 \sec^2\theta \tan^2 \theta}{(1 +\tan^3\theta)^2}d\theta =\frac{9a^2}{6}\int_0^{\pi/2}\frac{d \tan^3\theta }{(1 +\tan^3\theta)^2}$ $=-\frac{3a^2}{2}\frac{1}{1+\tan^3\theta}\bigg|_0^{\pi/2}=\frac{3a^2}{2}.$ In this case $a=\frac{1729}{3},$ so the area $A=\frac{1729^2}{6},$ and, therefore, the answer is $1735.$

You can also get the solution by usage of the formula for the cubic equation in y. This brings $y_{1}(x), y_{2}(x)$ , i.e. the two curves including the area wanted. After determination of maximum x to 914.872 by setting $y_{1}(x) = y_{2}(x)$ you have the integration range to numerically calculate the integral of the area v between the two curves delivering v = 498240.1666 => $6v = 1729^{2}$ .