[size=85]We construct triangle [/size][size=85] [math]ABC[/math] and Napolen cubic for this triangle. After that we take point [math]M[/math] on the cubic and then we cojugate that point isogonal we get point [math]N[/math] which is also on the Napoleon cubic. And by using GeoGebra we conclude that Napoleon cubic is isogonal transform of itself.[/size]

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

[size=85]Here we again substitude [math]x[/math], [math]y[/math] and [math]z[/math] with [math]a^2zy[/math], [math]b^2xz[/math] and [math]z^2xy[/math].[br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\right] x\left(c^2y^2-b^2z^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\right]a^2zy\left(c^2{(c^2xz)}^2-b^2(z^2xy)^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\right]a^2zy\left(c^2b^4x^2z^2-b^2z^4x^2y^2\right)=0[/math][br][math]\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\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[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\right]x\left(b^2z^2-c^2y^2\right)=0[/math][br][math]a^2b^2c^2xyz\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\right]x\left[-(c^2y^2-b^2z^2\right)]=0[/math][br][math]-a^2b^2c^2xyz\sum_{cyclic}\left[a^2\left(b^2+c^2\right)-\left(b^2-c^2\right)\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. Napoleon cubic is isogonal transform of itself.[/size]