This is the problem that was given to me. After going through google and looking in my book (Stewart Calculus) I am still stumped on this because it is not linear. My next instinct would be to check the derivatives instead of the actual curve.
Here's the parametric curve:
$ x = t^3 -3t$; $ y = t^3 - 3t^2$
t is all real numbers
You need to find $a,b: x(a)-x(b)=(a-b)(a^2+ab+b^2-3)=(a-b)(a^2+ba+b^2-3a-3b)=y(a)-y(b)=0$ this implies by substituting the first equation to the second one (since $a\ne{}b$) $3a+3b=3 \implies a+b=1$ thus $a^2+a-a^2+a^2-2a+1-3=0 \implies a^2-a-2=0 \implies a=-1$ or $a=2$ so your points are $t_1=-1,t_2=2$.
$t^3 -3t = u^3 -3u$ iff $t^3-u^3=3(t-u)$
$t^3 - 3t^2 = u^3 - 3u^2$ iff $t^3-u^3=3(t^2-u^2)$
Hence $3(t-u)=3(t^2-u^2)$ and so $t=u$ or $t+u=1$.
