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Note: (written November 6 2016) I'm glad you found this essay I wrote long ago. It's archived here on my website, together with my other essays from a decade ago. If you want to see my more recent content, my blog is the place to find it. Solving the Cubicby Walter VanniniAn Example Simple Solution Creative Solution Another Example Really Simple Solution Using The Sledgehammer Final Example No Simple Solution Cardano To The Rescue The Quadratic Equation The Cubic Equation Nomenclature Feedback IntroductionI've always thought that the explanations I've read of Cardano's solution to the general cubic were unnecessarily complicated. Whenever I ask someone who has been exposed to the method to explain it, the usual answer is that they don't remember how it went. This is my attempt to explain it in such a way that it has a better chance of being remembered. If nothing else, I'll remember it better! Since most approaches get very general very quickly, I'll do something else instead. I'll start with a progression of carefully chosen concrete examples. Then I'll generalize. The examples I'm going to use are: x^{3}  3x + 2 = 0x^{3}  3x = 0 x^{3}  3x + 1 = 0 Along the way I'll look at the quadratic equation and completing the square, describing that technique in a way that connects with solving the cubic. The first example won't need complex numbers, but soon after they'll be heavily used. If you're unfamiliar with complex numbers you're welcome to try and read this essay, but it will soon become really hard to follow. You've been warned! An ExampleThe example I like to start off with is x^{3}  3x + 2 = 0. Simple SolutionThe standard way to solve this is to try a few simple values: 1, 1, 2, 2. It turns out that 1 is a solution! That means (x1) is a factor of x^{3}  3x + 2 = 0. Polynomial division gives that x^{3}3x+2 can be factored as (x1)(x^{2} + x  2). This tells us that the solutions of x^{3}  3x + 2 = 0 are given by the solutions of x1=0 and x^{2}+x2=0. Solving for the quadratic factor can be done in many ways. In this case there is an obvious factorization x^{2} + x  2 = (x1)(x+2). So, x^{3}  3x + 2 = (x1)^{2}(x+2). The roots are 2 and 1 (with multiplicity 2). The only problem with this solution technique is that it depends on somehow intuiting a linear factor. When you can do it, you're in business. When you can't, you're stuck. Cardano's technique is more work, but it will always get you the answer. Creative SolutionHere's how this example is solved using Cardano's technique. The crucial creative step (in my opinion) is to replace x with (w+1/w), and to solve for w. This leads to a sixth order equation in w, BUT, it's a sixth order equation that's easy to solve. Here are the details: x^{3}  3 x + 2 =(w+1/w)^{3}  3(w+1/w) + 2 = w^{3} + 3 w^{2} (1/w) + 3 w (1/w)^{2} + (1/w)^{3}  3 (w+1/w) + 2 = w^{3} + 3 w + 3 (1/w) + (1/w^{3})  3 (w+1/w) + 2 = w^{3} + (1/w^{3}) + 2. So x^{3}3x+2=0 becomes w^{3}+1/w^{3}+2 = 0 This is a sixth order equation in w. The trick to solving it is to realize that it's really a quadratic in w^{3}. Replacing w^{3} with z, we get: z + 1/z + 2 = 0 ie z^{2} + 1 + 2z = 0. Solving the quadratic using whatever technique you like (I like z^{2} + 1 + 2z = (z+1)^{2} ) you get z = 1. So, w^{3} = 1. A solution to this is w = 1, which gives the solution x = w + 1/w = 1+1/1 = 2. Polynomial division now gives x^{3}3x+2 can be factored as (x+2)(x^{2}  2x + 1). x^{2}  2x + 1 = (x1)^{2}, so we have that the solutions are x = 2 and 1 (with multiplicity 2). If you know something about complex numbers, you might have noticed that I could have used other solutions to w^{3} = 1. The three solutions are: w = 1, 1/2 + i √3/2, 1/2  i √3/2 Choosing w = 1/2 + i √3/2, we get the solution x = w + 1/w = (1/2 + i √3/2) + (1/2  i √3/2) = (1/2)+(1/2) = 1 Something to notice here is that we managed to get a real solution to the cubic by using complex numbers! Also, in this particular example, using complex numbers is optional. By choosing the w= 1 solution to w^{3}= 1 we successfully avoided dealing with them. This doesn't always happen! When I wrote that this example was carefully chosen, I meant it. Another ExampleTo warm up a little more, consider x^{3}  3x = 0. Really Simple SolutionThis is very easy to solve without Cardano's technique. Since there's no constant term, x is a factor. x^{3}3x can be factored as x(x^{2}  3). This tells us that the solutions are x = 0, √3, √3. Using The SledgehammerReplace x with (w+1/w) x^{3}  3 x =(w+1/w)^{3}  3(w+1/w) = w^{3} + 3 w^{2} (1/w) + 3 w (1/w)^{2} + (1/w)^{3}  3 (w+1/w) = w^{3} + 3 w + 3 (1/w) + (1/w^{3})  3 (w+1/w) = w^{3} + (1/w^{3}). Replacing w^{3} with z, we get: z + 1/z = 0 z^{2} = 1 There's no escaping those complex numbers this time! z = i, i. Going with the i solution, we now need to solve w^{3} = i Well, one of the solutions is w = i. This gives us x = w+1/w = i + 1/i = i + i = 0. This tells us that (x0) is a factor of x^{3}3x, and the rest is routine: x^{3}3x = x(x^{2}3) giving solutions x = 0, √3, √3. Final ExampleNow consider x^{3}  3x + 1 = 0. No Simple SolutionAs far as I know, there is no simple way to solve this. Cardano To The RescueReplace x with (w+1/w) x^{3}  3 x + 1 =(w+1/w)^{3}  3(w+1/w) + 1 = w^{3} + 3 w^{2} (1/w) + 3 w (1/w)^{2} + (1/w)^{3}  3 (w+1/w) + 1 = w^{3} + 3 w + 3 (1/w) + (1/w^{3})  3 (w+1/w) + 1= w^{3} + (1/w^{3}) + 1. Replacing w^{3} with z, we get: z + 1/z + 1 = 0 z^{2} + 1 + z = 0 z = 1/2 + i √3/2, 1/2  i √3/2 Going with the 1/2 + i √3/2 solution, we now need to solve w^{3} = 1/2 + i √3/2 ie w^{3} = cos(120°) + i sin(120°) This means the solutions are: w = cos(40°) + i sin(40°), cos(160°) + i sin(160°), cos(280°) + i sin(280°) These give us x = w+1/w = 2cos(40°), 2 cos(160°), 2cos(280°) The Quadratic EquationThe general quadratic equation is: a x^{2} + b x + c = 0. The idea is to reduce it to another quadratic y^{2} = T. We know how to solve this: y= √T, −√T. Note: if a,b,c are real in the general equation, then T will be real in the reduced equation. I'm emphasizing this because the corresponding statement for the cubic equation is NOT true. The usual steps are: Divide by a to get an equation of the form x^{2} + B x + C = 0 (here B = b/a, C = c/a). Now replace x with y+k, where k is chosen so that the equation is of the form y^{2} + D = 0. This is done by choosing k to satisfy 2k = B, ie k = B/2. Taking T = D, we now have y^{2} = T. We're done! The Cubic EquationThe general cubic equation is: a x^{3} + b x^{2} + c x + d = 0 The idea is to reduce it to another cubic w^{3} = T. We know how to solve this. Note: even if a,b,c,d are real in the general equation, that does NOT mean that T will be real. It could be any complex value. There isn't that much more to it. All the important steps have been covered in the examples. The other ideas are present in the quadratic case. Here's what you do: Note that the first two steps have direct analogues in the quadratic case. Firstly divide by a to get an equation of the form x^{3} + B x^{2} + C x + D = 0. Now replace x with y+k, where k is chosen so that the equation is of the form y^{3} + p y + q = 0.
This is done by choosing k to satisfy 3k = −B, ie k = −B ⁄ 3. Now replace y with w+m/w, where m is chosen so that the w and 1/w terms disappear. This is done by choosing m to satisfy p = 3m, ie m =  p/3. This is why I chose p = 3 in the above examples. We now have a sixth order eqation in w of the form w^{3} + R(1/w^{3}) + S = 0. (or, if you prefer w^{6} + S w^{3} + R = 0) Taking z = w^{3}, we have a quadratic with 2 solutions, both of which can be complex. Pick one. Call it T. The general cubic has been reduced to the cubic w^{3} = T Solve this and reverse the above substitutions. NomenclatureThe substitution x = w+m/w is sometimes called Vieta's substitution. A sixth order equation of the form w^{6} + S w^{3} + R = 0 is sometimes called a triquadratic equation A third order expression of the form y^{3} + p y + q is sometimes called a depressed cubic FeedbackIf you have corrections, additions, modifications, etc please let me know mailto:walterv@gbbservices.com
