Answer
See explanations.
Work Step by Step
Step 1. Given $y=x^3$, we have $y'=3x^2$ which gives the slope of tangent lines to the curve.
Step 2. Given a point $(a, a^3)$, the slope of the tangent line is $m_1=3a^2$ and the line equation is $y-a^3=3a^2(x-a)$ or $y=3a^2x-2a^3$
Step 3. Find the other intersection between the above tangent line and the original curve. $x^3=3a^2x-2a^3$ or $x^3-3a^2x+2a^3=0$, which gives $x^3-a^2x-2a^2_2a^3=0$ or $x(x^2-a^2)-2a^2(x-a)=0$ or $(x-a)(x(x+a)-2a^2)=0$. Discard the solution of $x=a$, we have $x^2+ax-2a^2=0$
Step 4. We can factor the above equation as $(x-a)(x+2a)=0$; again, discarding the solution $x=a$, we get $x=-2a$ and $y=x^3=-8a^3$; thus the other intersection point is $(-2a, -8a^3)$
Step 5. The slope of the tangent line from the above intersection point is $m_2=3(-2a)^2=12a^2=4m_1$
Note: we can use synthetic division by $x-a$ twice to get the same results.
