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

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

Submitted by sudra on Fri, 03/14/2008 - 10:00

Forums:

Prove that the roots of the cubic equation x3-3a2x-2a3cos3A=0,

Are 3aCosA, 2aCos (120+_A)

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## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

> 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

## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

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

## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

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

## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

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).

## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

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

## Re: Prove that the roots of the cubic equation x3-3a2x-2a3co...

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

Cam