Fraction DecompositionI have the following problem:Suppose x + y + z = 0. Show that x 5 + y 5 + z 5 5 = x 3 + y 3 + z 3 3 × x 2 + y 2 + z 2 2 and x 7 + y 7 + z 7 7 = x 2 + y 2 + z 2 2 × x 5 + y 5 + z 5 5 I thought this was a fun problem to tackle, but I've been stuck on this for hours, and I can't seem to get anywhere. I've tried expanding, but no matter what I do, I just get a huge mess.I also cannot find how to incorporate the initial assumption ( x + y + z = 0) anywhere. Have I missed something?

Question

Fraction DecompositionI have the following problem:Suppose   x  +  y  +  z  =  0. Show that                    x        5            +              y        5            +              z        5              5    =                    x        3            +              y        3            +              z        3              3    ×                    x        2            +              y        2            +              z        2              2  and                    x        7            +              y        7            +              z        7              7    =                    x        2            +              y        2            +              z        2              2    ×                    x        5            +              y        5            +              z        5              5  I thought this was a fun problem to tackle, but I&#039;ve been stuck on this for hours, and I can&#039;t seem to get anywhere. I&#039;ve tried expanding, but no matter what I do, I just get a huge mess.I also cannot find how to incorporate the initial assumption (  x  +  y  +  z  =  0) anywhere. Have I missed something?

Jaylynn Huffman · Accepted Answer

You&#039;ve surely been trying to express one of the unknowns using the others and plugging in. Try to preserve the symmetry between them instead:  (  x  +  y  +  z      )    3    =      x    3    +      y    3    +      z    3    +  3      x    2    y  +  3      x    2    z  +  3      y    2    x  +  3      y    2    z  +  3      z    2    x  +  3      z    2    y  +  6  x  y  zBut we know that the left hand side is a zero and, moreover, wherever something like   x  +  z appears we can replace that by a   −  y (again, strictly keeping exchange symmetry). So the same equation can be written as  0  =      x    3    +      y    3    +      z    3    +  3      x    2    (  −  x  )  +  3      y    2    (  −  y  )  +  3      z    2    (  −  z  )  +  6  x  y  zand even better than that!                    0                            =        −        2        (                  x          3                +                  y          3                +                  z          3                )        +        6        x        y        z                            x        y        z                            =                                            x              3                        +                          y              3                        +                          z              3                                3                .            Surely replacing the right hand side by the left in your problem will make the multiplications simpler! Try to proceed from here :-)Hint: apart from       S          111        =  x  y  z and       S    k    =      x    k    +      y    k    +      z    k  , among which you have a first relation here, you&#039;ll need other symmetric forms like       S          11        =  x  y  +  x  z  +  y  zUpdate: I have been able to derive both your equations this way so it&#039;s guaranteed to work. It just takes some time. There&#039;s a whole theory about monomial and power-sum symmetric polynomials but I didn&#039;t want to go into that.Starting from       S    1    =  0, you can derive things like                              S          2                                    =        −        2                  S                      11                                                        S          3                                    =        3                  S                      111                                                        S          4                                    =                  1          2                          S          2          2                    and also some useful observations like                              S                      k            l                                              =                  S          k                          S          l                −                  S                      k            +            l                                                        S                      (            a            +            c            )            (            b            +            c            )            c                                              =                  S                      c            c            c                                    S                      a            b                          =                  S                      111                    c                          S                      a            b                              With these you&#039;re one little theorem away from       S    5        /    5  =      S    3        S    2        /    6 and       S    7        /    7  =      S    5        S    2        /    10

Aleah Booth · Accepted Answer

The first identity does not involve much algebra, if you write   z  =  −  (  x  +  y  )      z    5    +  (      x    5    +      y    5    )  =  −  5      x    4    y  −  10      x    3        y    2    −  10      x    2        y    3    −  5  x      y    4  where the   (  −      x    5    −      y    5    ) cancels out in the binomial exapnsion. A similar thing happends in the cubes term.So the identity reads  −      x    4    y  −  2      x    3        y    2    −  2      x    2        y    3    −  x      y    4    =  (  −      x    2    y  −  x      y    2    )            (              x        2            +              y        2            +              x        2            +      2      x      y      +              y        2            )        2  and of the six terms on the right, two pairs add together, and you get the left side.

Answered question

Answer & Explanation

New Questions in Pre-Algebra