Show that C is not simple by finding the point where the curve intersects itself

Question

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

Answer 1

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 $t_1=-1,t_2=2$.

Answer 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 $t+u=1$.