Fork me on GitHub
Math for the people, by the people.

User login

Prove that the roots of the cubic equation x3-3a2x-2a3cos3A=...

Primary tabs

Prove that the roots of the cubic equation x3-3a2x-2a3cos3A=...

Prove that the roots of the cubic equation x3-3a2x-2a3cos3A=0,
Are 3aCosA, 2aCos (120+_A)


> Just plug those quantities into your polynomial and show
> that you get zero. You'll probably need some double/triple
> angle identities.
>
> Cam

That's correct! Identity cos(3A) = 4cos^3(A) - 3cos(A) is susceptible to raising polynomial equations on the form
[2a.cos(A)]^3 - 3(2cosA) - 2a^3.cos(3A) = 0,
where x = 2a.cos(A) is a root of that equation. The remaining two one are found out by solving the resultant quadratic equation once the original one is reduced by using Ruffini or long division, for instance.
perucho

The intermediate term is -3[2a.cos(A)], not -3(2cosA). Excuse.

thank you clear how to proceeds.but the middle term is 3a^2X,like last term 2a^3cos(3A)

Right! sudra. Well done! In my correction post (the second one) I commited again a mistake (well, that is normal in me!). I glad you got it, and don't worry because all is okey. Let me show you.
cos(3A) = 4cos^3(A) - 3cos(A)
Multiplying by 2a^3 we have
2a^3cos(3A) = 8a^3cos^3(A) - 6a^3cos(A).
Now meet cosine powers with like "a" powers on RHS, i.e.
2a^3cos(3A) = [2a.cos(A)]^3 - 3a^2[2a.cos(A)],
and put now x = 2a.cos(A) to get
2a^3cos(3A) = x^3 - 3a^2x.
Sorry sudra.
perucho
PS. By the way sudra, roots of the equation also may be expressed as
x_1 = 2a.cos(A), x_2 = -2a.cos(60-A), x_3 = -2a.cos(60+A),
since
cos(120+A) = -cos(60-A), and cos(120-A) = -cos(60+A).

sudra, what are you putting there, son?
Please write powers x3 as x^3, a2 as a^2, etc. Also, cos3A is cos^3(A) or cos(3A)? So writes again your equation.
Ah, by the way, you have "eaten" one out of three real roots.
perucho

Just plug those quantities into your polynomial and show that you get zero. You'll probably need some double/triple angle identities.

Cam

Subscribe to Comments for "Prove that the roots of the cubic equation x3-3a2x-2a3cos3A=..."