Cubic equations

Recall that solving the quadratic equation,

x^2+ax+b = 0


x = \frac{1}{2}\,(-a \pm \sqrt{y}\,)

where y is the linear,

y = a^2-4b

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



x =\frac{1}{3}\,(-a+y_1^{1/3}+y_2^{1/3})

where the y_i are now the two roots of the quadratic,

y^2+(2a^3-9ab+27c)y+(a^2-3b)^3 = 0

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,

x = \frac{1}{5}\,(y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5})

where the y_i 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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