- Thread starter
- #1

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