Let a , b , c be nonzero real numbers and let a 2 − b 2 = b c and b 2 − c 2 = c a. Prove that a 2 − c 2 = a b.The solution strategy given in the course was to scale the two given equations by s = 1 c , resulting in a 2 − b 2 = b c becoming a 2 − b 2 = b and b 2 − c 2 = c a becoming b 2 − 1 = a. c is basically being set to 1, but I don't understand the justification. Doesn't scaling by 1 c by definition not change the equations, since ( a / c ) 2 − ( b / c ) 2 = ( b / c ) ( c / c ) ⟺ a 2 − b 2 c 2 = b c c 2 ⟺ a 2 − b 2 = b cWhere is mistake?

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Let   a  ,  b  ,  c be nonzero real numbers and let       a    2    −      b    2    =  b  c and       b    2    −      c    2    =  c  a. Prove that       a    2    −      c    2    =  a  b.The solution strategy given in the course was to scale the two given equations by   s  =      1    c  , resulting in       a    2    −      b    2    =  b  c becoming       a    2    −      b    2    =  b and       b    2    −      c    2    =  c  a becoming       b    2    −  1  =  a.   c is basically being set to 1, but I don&#039;t understand the justification. Doesn&#039;t scaling by       1    c   by definition not change the equations, since   (  a      /    c      )    2    −  (  b      /    c      )    2    =  (  b      /    c  )  (  c      /    c  )  ⟺                              a          2                −                  b          2                            c        2              =                    b        c                    c        2              ⟺      a    2    −      b    2    =  b  cWhere is mistake?

Aurora Hutchinson · Accepted Answer

You can think of this scaling as dividing all equalities by       c    2  , which is valid because   c  ≠  0. This division basically reduces a 3-variables problem into a 2-variables problem, which should be more tractable.Let   A  =  a      /    c,   B  =  b      /    c and   C  =  c      /    c  =  1. Then, we have      a    2    −      b    2    =  b  c    ⟹        A    2    −      B    2    =  B  ;        b    2    −      c    2    =  c  a    ⟹        B    2    −  1  =  A  .It follows that      A    2    −  1  =  (      A    2    −      B    2    )  +  (      B    2    −  1  )  =  B  +  A  .Thus, it remains that we show   A  +  B  =  A  B. Note that  B  =      A    2    −      B    2    =      A    2    −  (  A  +  1  )  =      A    2    −  A  −  1so                    (i)                    A        B        =        A        +        B                ⟺                          A          3                −                  A          2                −        A        =                  A          2                −        1                ⟺                          A          3                −        2                  A          2                −        A        +        1        =        0.            To show that (i) holds, we use  1  +  A  =      B    2    =  (      A    2    −  A  −  1      )    2    =      A    4    −  2      A    3    −      A    2    +  2  A  +  1which implies                    (ii)                    0        =                  A          4                −        2                  A          3                −                  A          2                +        A        =        A        (                  A          3                −        2                  A          2                −        A        +        1        )        .            Now we are done: (ii) implies that (i) is true, therefore we indeed have   A  B  =  A  +  B.

Let a,b,c be nonzero real numbers and let a^2−b^2=bc and b^2−c^2=ca. Prove that a^2−c^2=ab.

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