[size=85]Here we construct a triangle [math]ABC[/math]. After that we construct Lucas cubic for this triangle and we take point [math]M[/math] on it then we conjugate [math]M[/math] isogonal and we get point [math]N[/math]. By using GeoGebra we get another cubic, the blue one one the figure. Lucas cubic is isogonal transfom of K172.[/size]

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

[size=85] Here we again sustitude [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(b^2+c^2-a^2\right)a^2yz\left(\left(b^2xz\right)^2-\left(c^2xy\right)^2\right)=0[/math][br][br][math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2yz\left(b^4x^2z^2-c^4x^2y^2\right)=0[/math][br][br][math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2x^2yz\left[-\left(b^4z^2-c^4y^2\right)\right]=0[/math][br][br][math]xyz\sum_{cyclic}a^2\left(b^2+c^2-a^2\right)x\left(b^4z^2-c^4y^2\right)=0[/math][br][br]But the barycentric equation of the isogonal transform cubic is [math]\sum_{cyclic}a^2S_ax\left(b^4z^2-c^4y^2\right)=0[/math] where [math]S_A=\frac{b^2+c^2-a^2}{2}[/math] and in order to get the same equation we have to multiply [math]\sum_{cyclic}\left(b^2+c^2-a^2\right)a^2x^2yz\left[-\left(b^4z^2-c^4y^2\right)\right]=0[/math] by [math]\frac{2}{2}[/math] and the result is-[br][br][math]2xyz\sum_{cyclic}a^2\frac{\left(b^2+c^2-a^2\right)}{2}x\left(b^4z^2-c^4y^2\right)=0[/math] and this is the same equation as that of the isogonal transform cubic which means that Lucas cubic is isogonal transform of K172.[/size][br][br][br]