## Cubic equations

Recall that solving the quadratic equation,

$x^2+ax+b = 0$

gives,

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

where y is the linear,

$y = a^2-4b$

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

$x^3+ax^2+bx+c=0$

then,

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

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.