Let $a$ and $t$ be positive real numbers. Let $\ell$ be the tangent to the graph $C$ of $y=ax^3$ at a point $P$ $(t,at^3)$, and let $Q$ be the point at which $\ell$ intersects the curve $C$ again. Further, let $p$ be the line passing though the point $P$ parallel to the $x$-axis; let $q$ be the line passing through the point $Q$ parallel to the $y$-axis; let $R$ be the point of intersection of $p$ and $q$. $x$-coordinate of $Q$ is in the form $-Et$. Find $E$.
My attempt: We obtained the equation of the tangent $\ell$ from the slope $y'=2at^3$ at point $P$, then
$y=3at^2x-2at^3$
To find $x$-coordinate, we know that $Q$ is a point at which $\ell$ intersects the curve $C$, then
Equation of curve C = Equation of tangent line
$ax^3=3at^2x-2at^3$
$x^3-3t^2x+2t^3=0$
Now I do not know how I could continue from here, could you please hint me to the right direction?
You have your cubic equation $x^3 - 3t^2x+2t^3 = 0$, and you know that $x = t$ is a solution, because the line intersects the curve there. In fact, you know that $x = t$ is a double solution, because the line is tangent to the curve there.
Which is to say that $(x-t)^2 = x^2 - 2tx + t^2$ is a factor of $x^3-3t^2x+2t^3$. So you can perform the polynomial division $(x^3-3t^2x+2t^3):(x^2-2tx+t^2)$ to get the final factor of $x^3-3t^2x+2t^3$, which also gives you the final root of the cubic equation.
Or you could use Vieta's formulas: We see that the product of the roots is $-2t^3$ (the constant term, with a negative sign because we have an odd degree polynomial), and we know that two of the roots are $t$ and $t$. That should make it rather obvious what the third root is.