(also, that's an expression, not an equation! an equation is a statement "[stuff] = [other stuff]", and it can be true or false. an expression doesn't have an = sign -- it can't be true or false.)
(also yes, that ^. we know what you meant but technically exponents come before the negative sign)
you know that there are two values of t that make it 0. but the question asked for values of x -- t was just something we made up to make things easier!
so, for any value of t, how many corresponding values of x are there?
and as you mentioned before, you can do this more generally - no matter what value you have for t, you can always take the cube root and get a value for x
the whole point of this is that you're finding where t³-3t+1=0, and then trying to see if the results you get from that give you anything for the original
Well, it's almost done (assuming IVT is allowed) - we've got t^3-3t+1=0 has two or zero positive roots, and we've got one positive root is in (0,1), so it has two positive roots. Then t=x^2, which has two roots per positive t, so the original equation has four roots.
At this point, we can't distinguish mathematics from puzzling.
But that's kinda quirky problem solving technique after all
(which works only until you get to basic calculus, where you can identify local minima/maxima and etc directly)
The calculus way: Identify t=x^2 substitution. Identify that local minima/maxima are at t=-1 and t=1, and the graph has rotational symmetry around t=0. Plug in the three values to guess the shape of the graph and derive that two roots are positive and one is negative. Plug in to t=x^2 to conclude that there are four real roots.