Find the real values x , y , z such that { x + y 2 + z 3 = 21 ( 1 ) y + z 2 + x 3 = 71 ( 2 ) z + x 2 + y 3 = 45 ( 3 )

Question

Find the real values   x  ,  y  ,  z such that      {                            x          +                      y            2                    +                      z            3                    =          21                    (          1          )                                      y          +                      z            2                    +                      x            3                    =          71                    (          2          )                                      z          +                      x            2                    +                      y            3                    =          45                    (          3          )

kejohananws · Accepted Answer

By substitution  x  =  −      z    3    −      y    2    +  21we come to the system  {      z    6    +  2      y    2        z    3    +      y    4    +      y    3    −  42      z    3    −  42      y    2    +  z  +  396  =  0  ,  −      z    9    −  3      y    2        z    6    −  3      y    4        z    3    −      y    6    +  63      z    6    +  126      y    2        z    3    +  63      y    4    −  1323      z    3    −  1323      y    2    +      z    2    +  y  +  9190  =  0  }in two variables   y  ,  z  .. The necessary and sufficient condition for   z to be a root of the reduced system is its resultant in   y equals zero. Consider a simple example of the system   {  x  y  −  3  =  0  ,  x  +  y  −  4  =  0  } having       y    2    −  4  y  +  3 as the resultant in   x. Computing the resultant of the reduced system in   y with help of Maple, we obtain      z      27    −  189      z      24    +  15869      z      21    −  270      z      19    −  770589      z      18    −  806      z      17    +  2751      z      16    +  23703246      z      15    +  82077      z      14    −  1652484      z      13    −  476609381      z      12    −  3301322      z      11    +  43400763      z      10    +  6247199406      z    9    +  64051684      z    8    −  614744566      z    7    −  51522303964      z    6    −  586660519      z    5    +  4368480127      z    4    +  244239132451      z    3    +  2045234869      z    2    −  12927999002  z  −  506350844104  .Its integer zeros may be only the divisors of   506350844104  =      2    3    ⋅      7    3    ⋅  22717  ⋅  8123. With help of Maple it is easy to determine that   z  =  2 is the only integer root of the discriminant. Substituting it in the reduced system, we obtain   {      y    4    +      y    3    −  26  ∗      y    2    +  126  =  0  ,  −      y    6    +  39  ∗      y    4    −  507  ∗      y    2    +  y  +  2130  =  0  }. Let us continue to find integer solutions. Factoring   126  =  2  ⋅      3    2    ⋅  7 and   2130  =  2  ⋅  3  ⋅  5  ⋅  71 and making substitutions in the last system, we find   y  =  3 is its unique integer root. At last,   x  =  21  −      3    2    −      2    3    =  4.

Find the real values x , y , z such that { <mtable columnalign="left lef

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