MHB Using Trigonometry for solving a Cubic equation

[cbarker1]

Dear Everyone,

I need some help for this general cubic equation.

\[ax^3+bx^2+cx+d=0\]. First divide the equation by a
\[x^3+b/ax^2+c/ax+d/a=0\]

Let x=y-b/(3a)
...

$$y^3+py+q=0$$
where $$p=c/a-(b/a)^2)/3;q=d/a-(b/a*c/a)/3+(b/a)^3/27$$
$$y=cos \theta$$
then i need to use the identity for $$cos(3\theta)= 4cos^3(\theta)-3cos(\theta) $$

and that is where need the help.
What to do next?

Thank you for your patiences
Cbarker1

 

[MarkFL]

You may find this article informative:

Cubic function - Wikipedia, the free encyclopedia

 

[cbarker1]

I read that article already. I am still confuse with the substitution.

 

[I like Serena]

I read that article already. I am still confuse with the substitution.

Did you try the substitution $y=2\sqrt{-\frac p 3}\cos\theta$?

 

[cbarker1]

Why Do the substitution works?

 

[I like Serena]

Why Do the substitution works?

It is exactly the substitution the wiki article describes.
A great mathematician thought it up.
So yes, the substitution works.