Three vertices of a triangle lie on $y^2=x^3$ 
If ${P(a,b),Q(c,d),R(m,n)}$ are the centroid, orthocenter, circumcenter respectively of a scalene triangle with  vertices on the curve $y^2=x^3$, then find the value of $$\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$$


I know that the centroid, orthocenter and circumcenter lie on the straight line and centroid divides orthocenter and circumcenter in ratio $2:1$. Thus
$$a=\frac{2m+c}{3},b=\frac{2n+d}{3}$$ As you can see I am not able to use  $y^2=x^3$.

I think there is some trick involved!
 A: $\begin{array}{} G=(a,\sqrt{a^3}) & H=(c,\sqrt{c^3}) & O=(m,-\sqrt{m^3}) \end{array}$
$\begin{array}{} \frac{x_{H}-x_{G}}{x_{G}-x_{O}}=2 & \frac{c-a}{a-m}=2 & a=\frac{2m+c}{3} \end{array}$
$\begin{array}{} \frac{y_{H}-y_{G}}{y_{G}-y_{O}}=2 & \frac{\sqrt{c^3}-\sqrt{a^3}}{\sqrt{a^3}+\sqrt{m^3}}=2  \end{array}$
$\frac{\sqrt{c^3}-\sqrt{(\frac{2m+c}{3})^3}}{\sqrt{(\frac{2m+c}{3})^3}+\sqrt{m^3}}=2$
One solution that checks the system is:
$m=\frac{1}{2}\left( 2+\sqrt{6+\sqrt[3]{9}}-\sqrt{12-\sqrt[3]{9}}+\frac{30}{\sqrt{6+\sqrt[3]{9}}}  \right) ·c$
$\begin{array}{} m=α·c & α≈0.158873884116  \end{array}$
numerically checking the value of the expression k
$k=\frac{a}{b}+\frac{c}{d}+\frac{m}{n}$
$k=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{c}}-\frac{1}{\sqrt{m}}$
$\begin{array}{} m=α·c & a=\frac{1+2α}{3}  \end{array}$
$k=\left(\frac{\sqrt{3}}{\sqrt{1+2α}}+1-\frac{1}{\sqrt{α}}   \right)·\frac{1}{\sqrt{c}} ≈\frac{-2·10^{-15}}{\sqrt{c}}$
Indicating that the value of k is zero
Proof:
$k=\left(\frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}}   \right)·\frac{1}{\sqrt{c}} =0$
$\begin{array}{} \text{Solving} & \frac{\sqrt{3}}{\sqrt{1+2β}}+1-\frac{1}{\sqrt{β}}=0 & β=0.158873884146 \end{array}$
β is equal to α which is the solution of the initial system.

