finding tangents where the curve meets itself by solving two equations with two variables

Question

I was asked to finn the equation of the tangents where the curve

$x=t^2 - t$

$y=t^3-3t+1$

meets itself. So first I need to know where the curve meets itself by finding two values of the parameter $t_0$ and $t_1$ such that

$t_{0}^2 - t_0=t_{1}^2 - t_1$

$t_{0}^3-3t_0+1=t_{1}^3-3t_1+1$

Fron the first one I solve for $t_0$ using the quadratic formula but thn substituting in the second one I get nothing. I appreciate any hint

Answer 1

Hint:

(I use $t=t_0$ and $s=t_1$). For the $x$ coordinate : $$ t^2-t=s^2-s \iff (s-t)(s+t-1)=0 $$ for $s\ne t$ this means $s=1-t$, so, for the $y$ coordinate : $$ t^3-3t+1=s^3-3s+1 \iff t^3-3t+1=(1-t)^3+3(1-t)+1 $$ solve for $t$ the equation: $$ t^3-3t+1-(1-t)^3-3(1-t)-1=0 \iff (2t-1)(t+1)(t-2)=0 $$ and test the solutions that really gives the double point.