Recall that solving the quadratic equation,
where y is the linear,
The solution to the cubic equation is simply a generalization. Given,
where the are now the two roots of the quadratic,
Notice that the constant term of its resolvent is a perfect cube. For prime degrees, if the irreducible equation is solvable, then it can be solved analogously. For depressed quintics (an easy linear transformation to make a = 0), then,
where the are now the four roots of its resolvent quartic. For non-binomial quintics, this quartic will have a constant term that is a perfect fifth power. And so on for other prime degrees.