The tangent to the curve $y=x-x^3$ at point P meets the curve again at point Q. One point of trisection lies on $y$ axis. What is the equation of locus of other point of trisection? I have tried finding the slope, distance and tangent equation but ended up with no results and a tedious work. Can someone help me in giving an insight to this problem and its solution?

Let $P(p,p-p^3)$. Since $y'=1-3x^2$, the equation of the tangent line at $P$ is given by $$y-(p-p^3)=(1-3p^2)(x-p)$$ We want to find $x\
ot=p$ such that $$(x-x^3)-(p-p^3)=(1-3p^2)(x-p),$$ i.e. $$x^3-3p^2x+2p^3=0$$ Since the LHS is divisible by $(x-p)^2$, we obtain $$(x-p)^2(x+2p)=0.$$ Hence, $Q(-2p, 8p^3-2p)$. I think that you can continue from here.

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For the trisection points $R_1,R_2$ of the line segment $PQ$ :

$R_1(x_1,y_1)$ : $$x_1=\frac{2\times p+1\times (-2p)}{3}=0,\quad y_1=\frac{2\times (p-p^3)+1\times (8p^3-2p)}{3}$$

$R_2(x_2,y_2)$ : $$x_2=\frac{1\times p+2\times (-2p)}{3},\quad y_2=\frac{1\times (p-p^3)+2\times (8p^3-2p)}{3}$$