Find the equation of cubic polynomial

[songoku]

Homework Statement

If α, β and γ are roots of cubic polynomial and:
αβγ = 6
α²+β²+γ²=20
α³+β³+γ³=121

Find the equation of cubic polynomial

Homework Equations

vieta

The Attempt at a Solution

The equation is in the form:
x³ - (α+β+γ)x² + (αβ + αγ + βγ) x - αβγ = 0

But I don't know how to find α+β+γ and αβ + αγ + βγ. Thanks

 

[tiny-tim]

If α, β and γ are roots of cubic polynomial …

oh come on songoku! …

(x-α)(x-β)(x-γ) ? ​

 

[Joffan]

Three equations, three unknowns... a bit of devil to solve easily though.

For the first equation in αβγ space, I'm seeing four hyperbolic surfaces in +++ and three +-- octants
The second equation a sphere of course - this tells me the magnitudes of αβγ are all less than √20 < 4.5
The third equation therefore limits me to the +++ case (121 is nearly 5^3)...

 

[rock.freak667]

But I don't know how to find α+β+γ and αβ + αγ + βγ. Thanks

You are given what α²+β²+γ² is. So what happens if you consider what (α+β+γ)² is? (expand it out and see what terms you have)

 

[I like Serena]

Hi songoku!

Related to Vieta's formulas are Newton's identities.
They work out the same as rock.freak667's suggestion.

From these you can get a relation between your unknown coefficients and the equations that you are given.
You won't find nice round numbers though.

 

[songoku]

There are some posts above that I don't actually understand but let me try

let : α+β+γ = p ; αβ + αγ + βγ = q

(α+β+γ)² = α² + β² + γ² + 2 (αβ + αγ + βγ)
p² = 20 + 2q
q = 1/2 (p² - 20)

(α+β+γ)³ = α³ + β³ + γ³ + 3(α+β+γ)(αβ + αγ + βγ) - 3 αβγ
p³ = 121 + 3pq - 18

subs. q to second equation results in cubic equation in terms of p, then by using calculator I got p = -6.7

Am I correct? How to find p manually?

Thanks

 

[tiny-tim]

hi songoku!

your method looks fine

there's no easy way to solve a cubic polynomial (unless you know the roots are integers)

 

[songoku]

hi songoku!

your method looks fine

there's no easy way to solve a cubic polynomial (unless you know the roots are integers)

hi tiny-tim

*sigh...* hope that the root will be integer when exam comes...

thanks a lot for the help