Cutting tangents

Let $A(x_{0}, x_{0}^{3})$ for $x_{0} > 0$ with the tangent line $y - x_{0}^{3} = (3x_{0}^{2})(x-x_{0}) \Rightarrow y = (3x_{0}^{2})x - 2x_{0}^{3}.$ If this tangent line is to insect the curve again at point B, then let $B(-kx_{0}, -k^{3}x_{0}^{3})$ for $k > 0$ and substitute it into the tangent line:

$-k^3x_{0}^{3} = (3x_{0}^{2})(-kx_{0}) - 2x_{0}^{3} \Rightarrow 0 = k^3 -3k - 2 = (k-2)(k+1)^2 \Rightarrow k = -1, 2.$

After discarding the negative root, the gradient at point B computes to:

$\frac{dy}{dx}|_{x = -2x_{0}} = 3(-2x_{0})^{2} = \boxed{12x_{0}^{2}}$

which is 4x the gradient at point A.