Sum of cubed roots x 1 3 + x 2 3 + x 3 3...

hryggcx

hryggcx

Answered

2022-07-11

Sum of cubed roots
x 1 3 + x 2 3 + x 3 3
x 1 4 + x 2 4 + x 3 4
where x 1 , x 2 , x 3 are the roots of
x 3 + 2 x 2 + 3 x + 4 = 0
using Viete's formulas.
I know that x 1 2 + x 2 2 + x 3 2 = 2, as I already calculated that, but I can't seem to get the cube of the roots. I've tried
( x 1 2 + x 2 2 + x 3 2 ) ( x 1 + x 2 + x 3 )
but that did work.

Answer & Explanation

Dobermann82

Dobermann82

Expert

2022-07-12Added 15 answers

If x 1 , x 2 , x 3 are the roots of x 3 + 2 x 2 + 3 x + 4 = 0 then
x 3 + 2 x 2 + 3 x + 4 = ( x x 1 ) ( x x 2 ) ( x x 3 )
= x 3 ( x 1 + x 2 + x 3 ) x 2 + ( x 1 x 2 + x 1 x 3 + x 2 x 3 ) x x 1 x 2 x 3 = x 3 e 1 x 2 + e 2 x e 3 .
So e 1 = 2, e 2 = 3 and e 3 = 4.
Now the trick is to express the power sums x 1 3 + x 2 3 = x 3 3 and x 1 4 + x 2 4 = x 3 4 in terms of the elementary symmetric polynomials { x 1 + x 2 + x 3 , x 1 x 2 + x 1 x 3 + x 2 x 3 , x 1 x 2 x 3 }
In the case of the fourth power sums you should get
x 1 4 + x 2 4 + x 3 4 = e 1 4 4 e 1 2 e 2 + 4 e 1 e 3 + 2 e 2 2 = 18
Esmeralda Lane

Esmeralda Lane

Expert

2022-07-13Added 7 answers

I think what you need is Newton's identities, in particular the section about their application to the roots of a polynomial.

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