If a b = b c = c d , prove that a d = a 5 + b 2 c 2 + a 3 c 2 b 4 c + d 4 + b 2 c d 2 What I've done so far; a b = b c = c d = k a = b k , b = c k , c = d k a = c k 2 , b = d k 2 a = d k 3 I tried substituting above values in the right hand side of the equation to get a d but couldn't.

Question

If       a    b    =      b    c    =      c    d  , prove that       a    d    =                              a          5                +                  b          2                          c          2                +                  a          3                          c          2                                      b          4                c        +                  d          4                +                  b          2                c                  d          2                    What I&#039;ve done so far;       a    b    =      b    c    =      c    d    =  k    a  =  b  k  ,  b  =  c  k  ,  c  =  d  k    a  =  c      k    2    ,  b  =  d      k    2      a  =  d      k    3  I tried substituting above values in the right hand side of the equation to get       a    d   but couldn&#039;t.

Miriam Payne · Accepted Answer

Having   a  =  d      k    3    ,  b  =  d      k    2    ,  c  =  d  k gives you                                                                        a                5                            +                              b                2                                            c                2                            +                              a                3                                            c                2                                                                    b                4                            c              +                              d                4                            +                              b                2                            c                              d                2                                                                        =                                                            d                5                                            k                                  15                                            +                              d                2                                            k                4                            ⋅                              d                2                                            k                2                            +                              d                3                                            k                9                            ⋅                              d                2                                            k                2                                                                    d                4                                            k                8                            ⋅              d              k              +                              d                4                            +                              d                2                                            k                4                            ⋅              d              k              ⋅                              d                2                                                                                      =                                            d                              k                                  15                                            +                              k                6                            +              d                              k                                  11                                                                    d                              k                9                            +              1              +              d                              k                5                                                                                      =                                                            k                6                            (              d                              k                9                            +              1              +              d                              k                5                            )                                      d                              k                9                            +              1              +              d                              k                5                                                                                      =                              k            6                                                            =                  |                          k          3                          |

tenes6x · Accepted Answer

You did fine. Repalicng your values in the term inside the radical, you have      a    5    +      b    2        c    2    +      a    3        c    2    =      d    5        k          15        +      d    5        k          11        +      d    4        k    6        b    4    c  +      d    4    +      b    2    c      d    2    =      d    5        k    9    +      d    5        k    5    +      d    4  thet is to say that the ratio is        k    6   , its square root is       k    3   which is       a    d

If a b </mfrac> = b c </mfrac> = c d </mfr

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