I'm trying to solve the equation below for x: ( x 3 − 4 ) 3 = ( ( x 2 + 4 ) 2 3 + 4 ) 2

Question

I&#039;m trying to solve the equation below for x:  (      x    3    −  4      )    3    =  (            (              x        2            +      4              )        2              3    +  4      )    2

Oswaldo Rosales · Accepted Answer

We first remove the bizarre cube roots from our equation. To do that, we consider   y  =            x      3        −    4   and   z  =                    x        2            +      4        3   where   y  ,  z  &amp;gt;  0. So, we can rewrite our equation as  (      y    2        )    3    =  (      z    2    +  4      )    2    ⇒      y    3    =      z    2    +  4which looks good.We also have the following systems of equations from our definition of y and z and from the above equation,      {                                        x            3                    −                      y            2                    =          4                                                  y            3                    −                      z            2                    =          4                                                  z            3                    −                      x            2                    =          4                        which are really beautiful.Now in our original equation, since RHS is a square, we have       x    3    −  4  ≥  0    ⟹    x  ≥      4    3  . Now, we assume   x  ≥  z. From the above systems of equations we have      {                                        x            3                    −                      z            3                    =                      y            2                    −                      x            2                                                            y            3                    −                      x            3                    =                      z            2                    −                      y            2                                  Now since we had   x  ≥  z, we have   z  ≥  y  ≥  x which is not possible unless we have       x    3    −  4  ≥  0    ⟹    x  ≥      4    3  . Now, we assume   x  ≥  z. From the above systems of equations we have      {                                        x            3                    −                      z            3                    =                      y            2                    −                      x            2                                                            y            3                    −                      x            3                    =                      z            2                    −                      y            2                                  Now since we had   x  ≥  z, we have   y  ≥  x and   z  ≥  yThis implies                               x          =          2                     which is the only solution to the problem.

I'm trying to solve the equation below for x: ( x 3 </msup> −

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