What is the formula for finding the roots of a cubic equation?

hint:  Look up the rational zeroes theorem, then look at synthetic division, and work from there

It's complicated.
https://en.wikipedia.org/wiki/Cubic_function#General_formula

tough huh!

Petra Bonfert-Taylor gives a lovely example in the first lecture of her introductory complex analysis MOOC at Coursera on how the study of certain cubic equations led to an understanding of the necessity of extending the number system into the complex plane. She mentions 15th/16th Century mathematicians Tartaglia, Del Ferro and Cardano and then discusses Bombelli's Problem, the solutions of \(x^3\ =\ 15x\ + 4\).

Edited December 8, 20196 yr by Clear Kets

 Here is my take on it:  If a and b are any numbers, then (a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3.  And 3ab(a+ b)= 3a^2b+ 3ab^2.  Subtracting, (a+ b)^3- 3ab(a+ b)= a^3+ b^3.  

Let x= a+ b, m= 3ab, and n= a^3+ b^3.  That equation becomes the reduced cubic,  x^3+ mx= n ("reduced" because there is no x^2 term).   So given a and b, we can construct a cubic equation that has a+ b as a solution.  What about the other way?  Given m and n, can we determine a and b and so x?  

Of course we can!  From 3ab= m, b= m/3a.  Then a^3+ b^3= a^3+ (m/3a)^3= n.  Multiplying by a^3, (a^3)^2+ (m/3)^3= na^3.  That is a quadratic equation in a^3: (a^3)^2- n(a^3)+ (m/3)^3= 0.  Using the quadratic formula, a^3= (n+/- sqrt(n^2- 4(m/3)^3)/2.   We can write that in a little nicer form as a^3= (n/2)+/- sqrt((n/2)^2- (m/3)^3).  From a^3+ b^3= n, we have b^3= n- a^3= (n/2)-/+ sqrt((n/2)^2- (m/3)^3).  The various combinations of "+" or "-" give the three roots of the cubic equation.