[size=85]Here again we construct a triangle and after that Darboux cubic for this triangle. Then we take point [math]M[/math] on the cubic and after that we cojugate this point isogonal, we get another point [math]N[/math] on the same cubic. By using GeoGebra we come to the conlusion that Darboux cubic is isogonal transform of itself.[/size]

[math]\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]x\left(c^2y^2-b^2z^2\right)=0[/math]

[size=85]Here we substitude [math]x[/math], [math]y[/math] and [math]z[/math] in the same way as that in the previous cubics[br][math]\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right] x\left(c^2y^2-b^2z^2\right)=0[/math][br][math]\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]a^2zy\left(c^2{(c^2xz)}^2-b^2(z^2xy)^2\right)=0[/math][br][math]\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]a^2zy\left(c^2b^4x^2z^2-b^2z^4x^2y^2\right)=0[/math][br][math]\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]a^2zy\left(c^2b^4x^2z^2-b^2z^4x^2y^2\right)=0[/math][br][math]a^2b^2c^2xyz\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]x\left(b^2z^2-c^2y^2\right)=0[/math][br][math]a^2b^2c^2xyz\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]x\left[-(c^2y^2-b^2z^2\right)]=0[/math][br][math]-a^2b^2c^2xyz\sum_{cyclic}\left[2a^2\left(b^2+c^2\right)+\left(b^2-c^2\right)^{^2}-3a^4\right]x\left(c^2y^2-b^2z^2\right)=0[/math][br]And again we have the same equation as that in the beginning. darboux cubic is isogonal transform of itself.[/size]