If a be the A.M. and x, y be the two G.M.'s between b and c, show that

$straight x cubed plus straight y cubed space equals space 2 abc$

Solution

Since a is the A.M. between b and c

∴ $space space straight a space equals space fraction numerator straight b plus straight c over denominator 2 end fraction$ ...(i)
Since x, y are two G.M.'s between b and c

∴          b, x, y, c are in G.P.
Let r be the common ratio
                       $straight x equals br comma space space straight y space equals space br squared comma space space straight c space equals space br cubed$
                 L.H.S. = $straight x cubed plus straight y cubed space equals space left parenthesis br right parenthesis cubed space plus space left parenthesis br squared right parenthesis cubed space equals space straight b cubed straight r cubed space plus space straight b cubed straight r to the power of 6$
                           $space space equals straight b squared left parenthesis br cubed right parenthesis space plus space straight b left parenthesis br cubed right parenthesis squared space equals space straight b squared straight c plus bc squared$                      $open curly brackets because space straight c space equals space br cubed close curly brackets$
                             = bc (b + c) = bc(2a)                                 {By using (i)}
                             = 2abc = R.H.S.

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Permutations And Combinations

If a be the A.M. and x, y be the two G.M.'s between b and c, show that

$straight x cubed plus straight y cubed space equals space 2 abc$

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