Is there a simpler approach to these system of equations?I recently came across the following system of equations: x + y + z = 1 x 2 + y 2 + z 2 = 2 x 3 + y 3 + z 3 = 3And I have two questions:One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set: x + y + z = 1 x 2 + y 2 + z 2 = 2 . . . x n + y n + z n = nTwo, is there a simpler approach for the prior solution set than substitution?

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Is there a simpler approach to these system of equations?I recently came across the following system of equations:  x  +  y  +  z  =  1        x    2    +      y    2    +      z    2    =  2        x    3    +      y    3    +      z    3    =  3And I have two questions:One, is there a way to prove or disprove whether there is a solution for this particular set of equations? Furthermore, is there a way to expand the proof for a more generalized set of equations, that is for this set:  x  +  y  +  z  =  1        x    2    +      y    2    +      z    2    =  2    .  .  .        x    n    +      y    n    +      z    n    =  nTwo, is there a simpler approach for the prior solution set than substitution?

Chloe Melendez · Accepted Answer

Hint:  (  x  +  y  +  z      )    2    =      x    2    +      y    2    +      z    2    +  2  (  x  y  +  y  z  +  z  x  )  (  x  +  y  +  z      )    3    =      x    3    +      y    3    +      z    3    +  3  (  x  y  +  y  z  +  z  x  )  (  x  +  y  +  z  )  −  3  x  y  zSo, we can find   x  +  y  +  z  ,  x  y  +  y  z  +  z  x  ,  x  y  zNow   x  ,  y  ,  z are the roots of      t    3    −  (  x  +  y  +  z  )      t    2    +  (  x  y  +  y  z  +  z  x  )  t  −  x  y  z  =  0I think the same pattern can be applied for the higher powers

enclinesbnnbk · Accepted Answer

We will make this into a cubic equation in terms of   z. We have

Is there a simpler approach to these system of equations? I recently came across the following syst

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