A way to determine the cubic formula

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 u3+v3u^3+v^3 and uvuv, while I find uu and vv directly, and I also determine an expression for the discriminant. Here it goes:

Let's start with the equation we want to solve: x3+bx2+cx+d=0x^3+bx^2+cx+d=0 where b,c,db,c,d are complex numbers. Let its roots be x1,x2,x3x_1,x_2,x_3. By Vieta's formulas we know that:

x1+x2+x3=bx1x2+x1x3+x2x3=cx1x2x3=dx_1+x_2+x_3=-b \\ x_1x_2+x_1x_3+x_2x_3=c \\ x_1x_2x_3=-d

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


Expand it to obtain something interesting:


Now, let m=x1x22+x2x32+x3x12m=x_1x_2^2+x_2x_3^2+x_3x_1^2 and n=x12x2+x22x3+x32x1n=x_1^2x_2+x_2^2x_3+x_3^2x_1. So, Δ=(mn)2\Delta=(m-n)^2. Our first objective is to find mm and nn in terms of the coefficients. The trick is first find m+nm+n and mnmn:

m+n=x1x22+x2x32+x3x12+x12x2+x22x3+x32x1m+n=x1x2(x1+x2)+x2x3(x2+x3)+x3x1(x3+x1)m+n=x1x2(x1+x2+x3)+x2x3(x1+x2+x3)+x3x1(x3+x1+x2)3x1x2x3m+n=(x1+x2+x3)(x1x2+x1x3+x2x3)3x1x2x3m+n=bc+3dm+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{#D61F06}{x_3})+x_2x_3(\color{#D61F06}{x_1}+x_2+x_3)+x_3x_1(x_3+x_1+\color{#D61F06}{x_2})-\color{#D61F06}{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=(x1x22+x2x32+x3x12)(x12x2+x22x3+x32x1)mn=(x1x2)3+(x1x3)3+(x2x3)3+3(x1x2x3)2+x1x2x3(x13+x23+x33)mn=(x1x2+x1x3+x2x3)33(x1+x2+x3)(x1x2+x1x3+x2x3)(x1x2x3)+3(x1x2x3)2+3(x1x2x3)2+x1x2x3((x1+x2+x3)33(x1+x2+x3)(x1x2+x1x3+x2x3)+3x1x2x3)mn=c33bcd+3d2+3d2d(b3+3bc3d)mn=c33bcd+3d2+3d2+b3d3bcd+3d2mn=c36bcd+9d2+b3dmn=(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{#3D99F6}{(x_1x_2)^3+(x_1x_3)^3+(x_2x_3)^3}+3(x_1x_2x_3)^2+x_1x_2x_3(\color{#D61F06}{x_1^3+x_2^3+x_3^3}) \\ mn=\color{#3D99F6}{(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{#D61F06}{(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 (mn)2=(m+n)24mn(m-n)^2=(m+n)^2-4mn we can find Δ\Delta, and hence mnm-n:

Δ=(bc+3d)24(c36bcd+9d2+b3d)Δ=b2c26bcd+9d24c3+24bcd36d24b3dΔ=b2c24c34b3d+18bcd27d2\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>nm>n, then:


Easily we can find mm and nn:

m=bc+3d+Δ2n=bc+3dΔ2\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 uu and vv:

u=x1+x2w2+x3wv=x1+x2w+x3w2u=x_1+x_2w^2+x_3w \\ v=x_1+x_2w+x_3w^2

Notice that uw=x1w+x2+x3w2uw=x_1w+x_2+x_3w^2 and vw=x1w+x2w2+x3vw=x_1w+x_2w^2+x_3, similarly uw2=x1w2+x2w+x3uw^2=x_1w^2+x_2w+x_3 and vw2=x1w2+x2+x3wvw^2=x_1w^2+x_2+x_3w. We can sum them in 9 different ways, but only 3 produce results independent of ww, they are:

u+v=2x1x2x3uw+vw2=x1+2x2x3uw2+vw=x1x2+2x3u+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-b to each equation and dividing by 3 does the work:

x1=u+vb3x2=uw+vw2b3x3=uw2+vwb3x_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 uu and vv. From uu, cube both sides:

u3=x13+x23+x33+3(x12x2w2+x1x22w+x22x3w2+x2x32w+x32x1w2+x3x12w)+6x1x2x3u3=(x1+x2+x3)33(x1+x2+x3)(x1x2+x1x3+x2x3)+3x1x2x3+3(nw2+mw)+6x1x2x3u3=b3+3bc3d+3(bc+3dΔ2w2+bc+3d+Δ2w)6du3=b3+3bc9d+(3bc+9d)(w2+w)+3Δ(ww2)2u^3=\color{#D61F06}{x_1^3+x_2^3+x_3^3}+3(\color{#20A900}{x_1^2x_2w^2}+\color{#3D99F6}{x_1x_2^2w}+\color{#20A900}{x_2^2x_3w^2}+\color{#3D99F6}{x_2x_3^2w}+\color{#20A900}{x_3^2x_1w^2}+\color{#3D99F6}{x_3x_1^2w})+6x_1x_2x_3\\ u^3=\color{#D61F06}{(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{#20A900}{nw^2}+\color{#3D99F6}{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 w2+w=1w^2+w=-1 and ww2=3iw-w^2=\sqrt{3}i, then:

u3=2b3+6bc18d+3bc9d+33Δi2u=2b3+9bc27d+27Δ23u^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 vv:


We obtained back uu and vv 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 uu and vv 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, 27Δ-27\Delta can be written as Δ124Δ23\Delta_1^2-4\Delta_2^3. Find Δ1\Delta_1 and Δ2\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.

Note by Alan Enrique Ontiveros Salazar
5 years, 9 months ago

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

Venkata Karthik Bandaru - 5 years, 9 months ago

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This is great. The formula for x1x_1 is already ginormous, I can't imagine how x2x_2 and x3x_3 will look like even after simplifying.

Now do the quartic formula ahahahaha

Pi Han Goh - 5 years, 9 months ago

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For x2x_2 and x3x_3 we just have to add some ww's. For the quartic formula there is a similar approach, let me try to do it without die trying.

Alan Enrique Ontiveros Salazar - 5 years, 9 months ago

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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.

Michael Mendrin - 5 years, 9 months ago

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@Alan Enrique Ontiveros Salazar

Great job!

Swapnil Das - 5 years, 9 months ago

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Nice work. Appreciate it.

Sachin Vishwakarma - 5 years, 8 months ago

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Whoa !! You PLAYED well !!

Akshat Sharda - 5 years, 9 months ago

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A great thanks. This is very helpful :)

Nihar Mahajan - 5 years, 9 months ago

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This is very cool!

Akshay Yadav - 5 years, 9 months ago

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i dont understand since i am still a highschool student

jonathan dapadap - 5 years, 9 months ago

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We all are high school students bro :P

Swapnil Das - 5 years, 9 months ago

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