Solve the classic third degree equation

Question

Find $x$ such that $$x^3-6x^2-3x+2=0.$$

Solution: Classical cubic resolution gives three real solutions $\boxed{2+\sqrt{20}\cos\frac{2k\pi+\arccos\sqrt\frac 45}3,k=0,1,2}$

(write $x=2+\sqrt{20}\cos t$)

I have heard that this was a method that came from Abel or Galois. Is it true or I have made a mistake?

Answer 1

First use Depressed cubic

$$y=x-2\iff x=y+2$$

$$0=(y+2)^3-6(y+2)^2-3(y+2)+2=y^3-15y-20$$

Use How would you find the roots of $x^3-3x-1 = 0$

Answer 2

This has nothing to do with either Abel or Galois. The trigonometric way of solving cubic equations, when they have three real roots, is due to Viète.

Answer 3

Wikipedia identifies François Viète (or Vieta) as having originally derived the trigonometric formula for cubic equations with three real roots.

The solution itself is correct.