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 \(u^3+v^3\) and \(uv\), while I find \(u\) and \(v\) directly, and I also determine an expression for the discriminant. Here it goes:

Let's start with the equation we want to solve: \(x^3+bx^2+cx+d=0\) where \(b,c,d\) are complex numbers. Let its roots be \(x_1,x_2,x_3\). By Vieta's formulas we know that:

\(x_1+x_2+x_3=-b \\ x_1x_2+x_1x_3+x_2x_3=c \\ x_1x_2x_3=-d\)

Then, let \(\Delta\) be the discriminant of the equation and \(w\) a primitive cube root of unity. Then, by the definition of discriminant we have:

\(\Delta=((x_1-x_2)(x_2-x_3)(x_3-x_1))^2\)

Expand it to obtain something interesting:

\(\Delta=((x_1x_2^2+x_2x_3^2+x_3x_1^2)-(x_1^2x_2+x_2^2x_3+x_3^2x_1))^2\)

Now, let \(m=x_1x_2^2+x_2x_3^2+x_3x_1^2\) and \(n=x_1^2x_2+x_2^2x_3+x_3^2x_1\). So, \(\Delta=(m-n)^2\). Our first objective is to find \(m\) and \(n\) in terms of the coefficients. The trick is first find \(m+n\) and \(mn\):

\(m+n=x_1x_2^2+x_2x_3^2+x_3x_1^2+x_1^2x_2+x_2^2x_3+x_3^2x_1\\ m+n=x_1x_2(x_1+x_2)+x_2x_3(x_2+x_3)+x_3x_1(x_3+x_1) \\ m+n=x_1x_2(x_1+x_2+\color{red}{x_3})+x_2x_3(\color{red}{x_1}+x_2+x_3)+x_3x_1(x_3+x_1+\color{red}{x_2})-\color{red}{3x_1x_2x_3} \\ m+n=(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)-3x_1x_2x_3 \\ \boxed{m+n=-bc+3d}\)

\(mn=(x_1x_2^2+x_2x_3^2+x_3x_1^2)(x_1^2x_2+x_2^2x_3+x_3^2x_1) \\ mn=\color{blue}{(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3}+3(x_1x_2x_3)^2+x_1x_2x_3(\color{red}{x_1^3+x_2^3+x_3^3}) \\ mn=\color{blue}{(x_1x_2+x_1x_3+x_2x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)(x_1x_2x_3)+3(x_1x_2x_3)^2}+3(x_1x_2x_3)^2+x_1x_2x_3(\color{red}{(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)+3x_1x_2x_3}) \\ mn=c^3-3bcd+3d^2+3d^2-d(-b^3+3bc-3d)\\ mn=c^3-3bcd+3d^2+3d^2+b^3d-3bcd+3d^2 \\ \boxed{mn=c^3-6bcd+9d^2+b^3d}\)

Now, with the identity \((m-n)^2=(m+n)^2-4mn\) we can find \(\Delta\), and hence \(m-n\):

\(\Delta=(-bc+3d)^2-4(c^3-6bcd+9d^2+b^3d)\\ \Delta=b^2c^2-6bcd+9d^2-4c^3+24bcd-36d^2-4b^3d \\ \boxed{\Delta=b^2c^2-4c^3-4b^3d+18bcd-27d^2}\)

WLOG assume that \(m>n\), then:

\(m-n=\sqrt{\Delta}\)

Easily we can find \(m\) and \(n\):

\[\boxed{m=\dfrac{-bc+3d+\sqrt{\Delta}}{2}} \quad \boxed{n=\dfrac{-bc+3d-\sqrt{\Delta}}{2}}\]

Now, our second objetive is try to find the roots in terms of the coefficients. Let's introduce \(u\) and \(v\):

\(u=x_1+x_2w^2+x_3w \\ v=x_1+x_2w+x_3w^2\)

Notice that \(uw=x_1w+x_2+x_3w^2\) and \(vw=x_1w+x_2w^2+x_3\), similarly \(uw^2=x_1w^2+x_2w+x_3\) and \(vw^2=x_1w^2+x_2+x_3w\). We can sum them in 9 different ways, but only 3 produce results independent of \(w\), they are:

\(u+v=2x_1-x_2-x_3 \\ uw+vw^2=-x_1+2x_2-x_3 \\ uw^2+vw=-x_1-x_2+2x_3\)

From here, see how we can obtain the roots. Yes, simply adding \(-b\) to each equation and dividing by 3 does the work:

\(x_1=\dfrac{u+v-b}{3} \\ x_2=\dfrac{uw+vw^2-b}{3} \\ x_3=\dfrac{uw^2+vw-b}{3}\)

Then, our last objective is to determine \(u\) and \(v\). From \(u\), cube both sides:

\(u^3=\color{red}{x_1^3+x_2^3+x_3^3}+3(\color{green}{x_1^2x_2w^2}+\color{blue}{x_1x_2^2w}+\color{green}{x_2^2x_3w^2}+\color{blue}{x_2x_3^2w}+\color{green}{x_3^2x_1w^2}+\color{blue}{x_3x_1^2w})+6x_1x_2x_3\\ u^3=\color{red}{(x_1+x_2+x_3)^3-3(x_1+x_2+x_3)(x_1x_2+x_1x_3+x_2x_3)+3x_1x_2x_3}+3(\color{green}{nw^2}+\color{blue}{mw})+6x_1x_2x_3 \\ u^3=-b^3+3bc-3d+3\left(\dfrac{-bc+3d-\sqrt{\Delta}}{2}w^2+\dfrac{-bc+3d+\sqrt{\Delta}}{2}w\right)-6d \\ u^3=-b^3+3bc-9d+\dfrac{(-3bc+9d)(w^2+w)+3\sqrt{\Delta}(w-w^2)}{2}\)

Remember that \(w^2+w=-1\) and \(w-w^2=\sqrt{3}i\), then:

\(u^3=\dfrac{-2b^3+6bc-18d+3bc-9d+3\sqrt{3}\sqrt{\Delta}i}{2} \\ \boxed{u=\sqrt[3]{\dfrac{-2b^3+9bc-27d+\sqrt{-27\Delta}}{2}}}\)

In a similar way we can obtain \(v\):

\(\boxed{v=\sqrt[3]{\dfrac{-2b^3+9bc-27d-\sqrt{-27\Delta}}{2}}}\)

We obtained back \(u\) and \(v\) 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 \(u\) and \(v\) to form the roots.

If we plug in everything, we obtain the messy formulas for the roots, this is one of them:

\(x_1=\dfrac{\sqrt[3]{\dfrac{-2b^3+9bc-27d+\sqrt{-27\Delta}}{2}}+\sqrt[3]{\dfrac{-2b^3+9bc-27d-\sqrt{-27\Delta}}{2}}-b}{3}\)

**Bonus question:** the expression inside the square root, \(-27\Delta\) can be written as \(\Delta_1^2-4\Delta_2^3\). Find \(\Delta_1\) and \(\Delta_2\).

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.

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## Comments

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TopNewestI am tempted to appreciate your patience in typing this out with \( \LaTeX \). Thank you !

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@Alan Enrique Ontiveros Salazar

Great job!

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I think the analogy would be that Your Approach vs Cardano's is like Chemistry vs Alchemy. Very nice!

This is the kind of thing that, although very long, I probably could still [eventually] remember how it could be done. Cardano's? I could never remember his way.

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This is great. The formula for \(x_1\) is already ginormous, I can't imagine how \(x_2\) and \(x_3\) will look like even after simplifying.

Now do the quartic formula ahahahaha

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For \(x_2\) and \(x_3\) we just have to add some \(w\)'s. For the quartic formula there is a similar approach, let me try to do it without die trying.

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Nice work. Appreciate it.

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This is very cool!

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A great thanks. This is very helpful :)

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Whoa !! You PLAYED well !!

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i dont understand since i am still a highschool student

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We all are high school students bro :P

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