Algebra

Let the root be $z$.

We have the equation: $z^3 = 1$

Subtracting 1 from both sides gives: $z^3 - 1 = 0$

Factoring out the left-hand side, we get: $(z-1)(z^2 + z + 1) = 0$

Oh, look... It's a quadratic that so happens to be in the form $ax^2 + bx + c = 0$

This means that the quadratic we are looking for is $z^2 + z + 1 = 0$, so the ordered pair is $\color{brown}\boxed{(1,1)}$