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 = 0using 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.

Question

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  =  0using Viete&#039;s formulas.I know that       x    1    2    +      x    2    2    +      x    3    2    =  −  2, as I already calculated that, but I can&#039;t seem to get the cube of the roots. I&#039;ve tried  (      x    1    2    +      x    2    2    +      x    3    2    )  (      x    1    +      x    2    +      x    3    )but that did work.

Dobermann82 · Accepted Answer

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 · Accepted Answer

I think what you need is Newton&#039;s identities, in particular the section about their application to the roots of a polynomial.

Sum of cubed roots <msubsup> x 1 3 </msubsup> + <msubsup> x 2 3

Answered question

Answer & Explanation

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