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?
 A: 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 $x\not=p$ 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.

For the trisection points $R_1,R_2$ of the line segment $PQ$ :
$R_1(x_1,y_1)$ : $$x_1=\frac{2\times p+1\times (-2p)}{3}=0,\quad y_1=\frac{2\times (p-p^3)+1\times (8p^3-2p)}{3}$$
$R_2(x_2,y_2)$ : $$x_2=\frac{1\times p+2\times (-2p)}{3},\quad y_2=\frac{1\times (p-p^3)+2\times (8p^3-2p)}{3}$$ 
A: Given $y=x-x^3$ and a tangent at point $P(a,a-a^3)$ also intersects the graph at a point $Q$.
The slope of the tangent line is $y=1-3x^2$, so the slope of the tangent line at $(a,a-a^3)$ on the graph is $m=1-3a^2$.
So the equation of the tangent line at that point is 
\begin{equation}
y-a+a^3=(1-3a^2)(x-a)
\end{equation}
or
\begin{equation}
y=(1-3a^2)x+2a^3
\end{equation}
Since $Q$ lies on the tangent line and also on the curve, the $x$ coordinate of $Q$ satisfies the equation
\begin{equation}
x-x^3=(1-3a^2)x+2a^3
\end{equation}
which can be re-written
\begin{equation}
x^3-3a^2x+2a^3=0
\end{equation}
Since $P$ also lies on the tangent line and the curve we know that $x=a$ is one solution of this equation. Thus we may factor out $x-a$ to obtain the equation
\begin{equation}
(x-a)(x^2+ax-2a^2)=0
\end{equation}
and the quadratic can also be factored to give
\begin{equation}
(x-a)^2(x+2a)=0
\end{equation}
We should not be surprised to see that $a$ is a double solution since that is where the tangent it. But the tangent line crosses the graph at $x=-2a$ so that is a single root of the polynomial.
The coordinates of $Q$ are $(-2a,-2a+8a^3)$.
