No slide rule required!

There must be $2$ different values of the parameter $t$ which yields the same $x$ , so first we solve the quadratic equation for $t$

${t}^{2}-t=x$

with solutions

$t_1=\dfrac{1}{2}\left( 1-\sqrt{1+4x} \right)$
$t_2=\dfrac{1}{2}\left( 1+\sqrt{1+4x} \right)$

We plug those into the equation for $y$

${t_1}^{3}-3t_1={t_2}^{3}-3t_2$

Simplifying, we end up with

$(x-2)\sqrt{1+4x}=0$

or $x=2$ , and we quickly find $y=2$ , so that the answer is $2+2=4$