How to solve an equation of third degree?

Question

I'm trying to solve this equation $x^3+bx^2+cx+d = 0$ and try to follow the steps from my notebook, but I barely understand my writing.

$(x+b/3)^3+bx^2+cx+d-xb^2/3-b^3/27$

$(x+b/3)^3+bx^2+x(c-b^2/3)+d-b^3/27$

Let $y= x+b/3$

$y^3+qy+r = 0$

I found out that $ q = (c-b^2/3)$ and $ r = d + (2b^3-3bc)/27.$ Is it correct ?

Let $y= \alpha + \beta $

$(\alpha+\beta)^3+q(\alpha+\beta)+r = 0$

$\alpha^3+3\alpha^2\beta+3\alpha\beta^3+\beta^3+q(\alpha+\beta)+r = 0$

$\alpha^3+\beta^3+(3\alpha\beta+q)(\alpha+\beta)+r = 0$

Then it says that $ \alpha^3+\beta^3 = -r $ and $\alpha\beta = -q/3 $. I can't figure out why.

Equation $t^2+rt-q^3/27=0 $ has the roots $ \alpha^3, \beta^3.$ I know this is from Vieta's formulas.

$ t_{1,2} = \frac {-r\pm\sqrt{r^2-4q^3/27}}{2}$

$\alpha_0 = \sqrt \alpha^3, \beta_0=\sqrt \beta^3$

$\alpha^3 = t_1$ has the roots $ \alpha_0,\epsilon\alpha_0,\epsilon^2\alpha_0.$

$\beta^3 = t_2 $ has the roots $ \beta_0,\epsilon\beta_0,\epsilon^2\beta_0.$

I'm lost here. I think that by $\epsilon$ he ment $\frac {cos2\pi}3 +i\frac {sin2\pi}3$. I don't understand why those are the roots.

The solutions of the equation $y^3+qy+r=0$ are $\alpha_0+\beta_0,\epsilon\alpha_0+\epsilon^2\beta_0,\epsilon^2\alpha_0+\epsilon\beta_0.$ Okay, but why ?

$ q = -1, r = 1.$ This makes no sense to me.

$\alpha_0 = \sqrt[3]{\frac {-1+\sqrt {1-4/27}}2} $

$\beta_0 = \sqrt[3]{\frac {-1-\sqrt {1-4/27}}2} $

$\alpha_0+\beta_0 \in R$

And that's all I have about cubic equations. I have the feeling that is uncomplete.

Answer 1

Hints.

1) Transform the equation to the form $x^3 + 3bx + c$. I see you already did it. In your case you have the form $y^3 + qy + r$ which is exactly the same.

2) Smartass trick: let $y = w - \frac{b}{w}$. This transforms the equation into $$w^3 + c - \frac{b^3}{w^3} = 0$$ which is a quadratic in $w^3$. So solve this

3) You now have a value for $w$.

Can you find then a root now?

Answer 2

The general strategy is to eliminate the quadratic term first, so as to have an equation in standard form $y^3+py+q=0$.

Then, Cardano's method replaces the single unknown $y$ by two unknowns $u$ and $v$ such that $y=u+v$. This introduces one degree of liberty, as physicists say. This allows to add a constraint, which will simplify the equation, namely eliminating the term with factor $u+v$.

Finally it comes down to determining two numbers $u$ and $v$, knowing the sum of their cubes, and their product. A problem, which in turn comes down to determining $u^3$ and $v^3$, knowing their sum and their product – a classical problem on quadatic equations.