Consider the system of equations in real numbers x , y , z satisfying x 4 x 2 + 1 = y 5 y 2 + 1 = z 6 z 2 + 1 and x + y + z = x y z . Find x , y , z ..I substituted x , y , z as tan ⁡ ( θ 1 ) , tan ⁡ ( θ 2 ) , tan ⁡ ( θ 3 ) , where θ 1 + θ 2 + θ 3 = 180 º .

Question

Consider the system of equations in real numbers      x  ,  y  ,  z    satisfying      x          4                               x          2                +        1                 =         y          5                               y          2                +        1                 =         z          6                               z          2                +        1            and      x  +  y  +  z     =     x  y  z   . Find      x  ,  y  ,  z     ..I substituted      x  ,  y  ,  z    as      tan  ⁡  (      θ    1    )     ,     tan  ⁡  (      θ    2    )     ,     tan  ⁡  (      θ    3    )   , where      θ    1       +         θ    2       +         θ    3       =     180  º        .

toriannucz · Accepted Answer

to any values   a  ,  b  ,  c instead of       1    4    ,      1    5    ,      1    6  If we put                    (1)                    ϕ        (        a        ,        b        ,        c        )        =        a        b        +        a        c        −        b        c        ,                 ψ        (        a        ,        b        ,        c        )        =        a        b        +        a        c        +        b        c            then LAcarguy’s method leads to the following explicit formulas for the solutions :                    (3)                    x        =                  ε                      ϕ            (                          a              2                        ,                          b              2                        ,                          c              2                        )                                    G          (          a          ,          b          ,          c          )                ,                 y        =                  ε                      ϕ            (                          b              2                        ,                          c              2                        ,                          a              2                        )                                    G          (          a          ,          b          ,          c          )                ,                 z        =                  ε                      ϕ            (                          c              2                        ,                          a              2                        ,                          b              2                        )                                    G          (          a          ,          b          ,          c          )                    where   ε is   +  1 or   −  1. In your initial question where   a  =      1    4    ,  b  =      1    5    ,  c  =      1    6  , one finds   G  =      7    921600  ,                    (4)                    x        =        ε                              7                    3                ,                 y        =        ε                              5                          7                                9                ,                 z        =        3        ε                  7

Ayaan Barr · Accepted Answer

Starting with             s      i      n      t        4    =            s      i      n      u        5    =            s      i      n      w        6  , then use the law of sines to get:       a    4    =      b    5    =      c    6  . So   a  =            2      c        3  ,   b  =            5      c        6  . So apply the law of cosines to have:       a    2    =      b    2    +      c    2    −  2  b  c  c  o  s  A. Substituting these values of   a and   b into the above equation we can solve for   c  o  s  A, and then   t  a  n  B, and   t  a  n  A, and   t  a  n  C.