If a is the A.M. of b and c and the two geometric means are `G_(1)` and `G_(2)`, then prove that `G_(1)^(3)+G_(2)^(3) = 2abc`

Explanation

Given, a=AM between b and c
`implies " " a=(b+c)/(2) implies 2a=b+c " " "………(i)"`
Again, `b,G_(1),G_(2),c` are in GP.
` therefore " " (G_(1))/(b)=(G_(2))/(G_(1))=(c)/(G_(2)) implies b=(G_(1)^(2))/(G_(2)),c=(G_(2)^(2))/(G_(1))`
and `G_(1)G_(2)=bc " " ".......(ii)"`
From Eqs. (i) and (ii),
`2a= (G_(1)^(2))/(G_(1))+(G_(2)^(2))/(G_(2))=(G_(1)^(3)+ G_(2)^(3))/(G_(1)G_(2))=(G_(1)^(3)+G_(2)^(3))/(bc)[therefore G_(1)G_(2)=bc]`
`implies G_(1)^(3)+G_(2)^(3)=2abc`