Answer
See below for detailed work.
Work Step by Step
$$y=x^3$$
1) Find the derivative of $y$: $$y'=3x^2$$
2) At any point $(a,a^3)$, which we would call point $A$, the slope of the tangent line is $$y'(a)=3a^2$$
Therefore, the equation of the tangent at $A$ is $$y-a^3=3a^2(x-a)$$ $$y-a^3=3a^2x-3a^3$$ $$y=3a^2x-2a^3$$
3) According to the claim, this tangent would, first, meet the curve again at a point $B$. We would try to find this point $B$.
$$x^3=3a^2x-2a^3$$ $$x^3-3a^2x+2a^3=0$$
Since we know $x=a$ is already a solution, we would try to create the $(x-a)$ in this equation: $$(x^3-a^2x)+(-2a^2x+2a^3)=0$$ $$x(x^2-a^2)-2a^2(x-a)=0$$ $$x(x-a)(x+a)-2a^2(x-a)=0$$ $$(x-a)\Big(x(x+a)-2a^2\Big)=0$$ $$(x-a)(x^2+ax-2a^2)=0$$ $$(x-a)\Big((x^2-ax)+(2ax-2a^2)\Big)=0$$ $$(x-a)\Big(x(x-a)+2a(x-a)\Big)=0$$ $$(x-a)^2(x+2a)=0$$
So either $x=a$ or $x=-2a$. $x=a$ already corresponds to $A$, so $x=-2a$ corresponds to $B$. In other words, $B(-2a,y)$
4) The claim continues that the slope here is 4 times the slope at $A$. We know the slope at $A$ is $y'(a)=3a^2$.
$$y'(-2a)=3\times(-2a)^2=3\times4a^2=12a=4(3a^2)$$
The slope at $B$ really is 4 times the slope at $A$.
