How to solve this sytem of equation: { ( a 2 + 4 + a ) ( b 2 + 1 + b ) = 1 27 a 6 + 8 b = a 3 + 2 I tried to analysis equation (1): a 2 + 4 + a = b 2 + 1 − b ⇔ a + b = b 2 + 1 − a 2 + 4 ⇔ 4 a 2 + 16 b 2 + 20 a b = 9And then, I don't know how to solve this.

Question

How to solve this sytem of equation:      {                            (                                    a              2                        +            4                    +          a          )          (                                    b              2                        +            1                    +          b          )          =          1                                      27                      a            6                    +          8          b          =                      a            3                    +          2                        I tried to analysis equation (1):            a      2        +    4    +  a  =            b      2        +    1    −  b  ⇔  a  +  b  =            b      2        +    1    −            a      2        +    4    ⇔  4      a    2    +  16      b    2    +  20  a  b  =  9And then, I don&#039;t know how to solve this.

Carlos Nicholson · Accepted Answer

you can solve the first equation for   b:  b  =  −  1  /  2              2                          a              2            +      3      +      2                                      a                  2                +        4            a                                        a                  2                +        4            +      a      plugging this in the second equation we get            27                          a              6                                      a                  2                +        4            +      27                          a              7            −                    a              3                                      a                  2                +        4            −                    a              4            −      8                                      a                  2                +        4            a      −      8                          a              2            −      2                                      a                  2                +        4            −      2            a      −      12                                        a                  2                +        4            +      a        =  0solving this for             a      2        +    4   and squaring again we get this polynomial:  729          a      12    −  54          a      9    −  270          a      7    −  107          a      6    +  10          a      4    +  4          a      3    +  16          a      2    +  20    a  −  32  =  0with only two real Solutions. this can be proven by using the Theorem of Sturm.

How to solve this sytem of equation: { <mtable columnalign="left left" rowspacing=".

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