Hi everyone, some time ago I was playing with Cardano's method to solve the general cubic equation. Then I discovered a new way to determine the same formula, altough is a little longer (also because I wanted to detail every step I do). It is very similar to the method of Lagrange, with some differences. That method finds and , while I find and directly, and I also determine an expression for the discriminant. Here it goes:
Let's start with the equation we want to solve: where are complex numbers. Let its roots be . By Vieta's formulas we know that:
Then, let be the discriminant of the equation and a primitive cube root of unity. Then, by the definition of discriminant we have:
Expand it to obtain something interesting:
Now, let and . So, . Our first objective is to find and in terms of the coefficients. The trick is first find and :
Now, with the identity we can find , and hence :
WLOG assume that , then:
Easily we can find and :
Now, our second objetive is try to find the roots in terms of the coefficients. Let's introduce and :
Notice that and , similarly and . We can sum them in 9 different ways, but only 3 produce results independent of , they are:
From here, see how we can obtain the roots. Yes, simply adding to each equation and dividing by 3 does the work:
Then, our last objective is to determine and . From , cube both sides:
Remember that and , then:
In a similar way we can obtain :
We obtained back and by taking cube root in both sides. We don't worry about the three possible cube roots, because we worked on that when we combined and to form the roots.
If we plug in everything, we obtain the messy formulas for the roots, this is one of them:
Bonus question: the expression inside the square root, can be written as . Find and .
I hope that you liked this note, it's my first note in Brilliant :D If you think that some steps could have been simplified, feel free to tell me.