Recall that solving the *quadratic equation*,

gives,

where y is the linear,

The solution to the *cubic equation* is simply a generalization. Given,

then,

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.

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