In FP1 we study how to get two more roots of a cubic equation if we are given one in advance. But life is not always so generous. How do we solve a cubic if we’re given no root in advance?

First, in our cubic equation

x^{3}+c_{1}x^{2}+c_{2}x+c_{3}=0

make a substitution z=(x−something) to change the equation to a simpler form

z^{3}−3pz−q=0

Now the sum of the roots is zero. So we can write the roots as

root 1 = a+b

root 2 = ωa+ω^{2}b

root 3 = ω^{2}a+ωb

if ω=cos(2π/3)+i.sin(2π/3), and so 1, ω, and ω^{2} are the three cube roots of 1.

1+ω+ω^{2}=0

so sum of the roots=0

sum of (root 1)×(root 2)+(root 1)×(root 3)+(root 2)×(root 3)=−3ab

product of the roots=a^{3}+b^{3}

∴ ab=p, or b=(p/a)

and a^{3}+b^{3}=q

so a^{3}+p/a^{3}=q

Let A=a^{3}, then

A+p/A=q, or A^{2}−qA+p=0

We can solve this because it is a quadratic in A; and then we can find a, then b, then the three roots.

Click here for more explanation by Tim Gowers

Click here for Vieta’s substitution – a similar method, but not using ω.

Click here for Lagrange’s method, which led directly into Abel’s and Galois’s work to prove that there is no similar algebraic “formula” for the quintic and into group theory, one of the central parts of modern mathematics.

Historically the problem of solving cubic equations is important also because even when all the three roots of the cubic are real, you get complex numbers in the working.

At first mathematicians used those complex numbers in the working without admitting they were actually numbers, rather as ancient Greeks and Romans did calculations with fractions without admitting fractions were numbers.