MHB Using Trigonometry for solving a Cubic equation

AI Thread Summary
The discussion centers on solving a cubic equation using trigonometric methods. The user seeks assistance after transforming the cubic equation into a simpler form and expressing it in terms of cosine. A suggested substitution is \(y=2\sqrt{-\frac{p}{3}}\cos\theta\), which aligns with methods outlined in a referenced Wikipedia article. Participants confirm the validity of this substitution and emphasize its historical mathematical significance. The conversation highlights the importance of understanding the substitution process for solving cubic equations.
cbarker1
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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
 
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I read that article already. I am still confuse with the substitution.
 
Cbarker1 said:
I read that article already. I am still confuse with the substitution.

Did you try the substitution $y=2\sqrt{-\frac p 3}\cos\theta$?
 
Why Do the substitution works?
 
Cbarker1 said:
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.
 
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