#a# i s The arithmetic mean of two positive numbers #b and c# . #G_1# and #G_2# are the geometric mean between the same positive numbers #b and c# so prove that #G_1^3+G_2^3#=#2abc# ?

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Answer 1 · 2018-08-09T11:50:59

Please see below.

Here,

#color(blue)("a is the AM of b and c " =>(b+c)/2=a=>b+c=2ato(1)#

Now, #color(red)(G_1 and G_2 " are the GM between b and c."#

So, #b,G_1,G_2,c " are in GP"#

Let #r " be the common ratio and b is the first term."#

So, #b, br, br^2,br^3 " are in GP"#

#i.e. color(blue)(G_1=br ,G_2=br^2 ,c=br^3......to(2)#

Let us take LHS.

#LHS=(G_1)^3+(G_2)^3#

#LHS=(br)^3+(br^2)^3#

#LHS=b^3r^3+b^3r^6#

#LHS=b^2color(blue)((br^3))+b(color(blue)(br^3))^2#

#LHS=b^2color(blue)((c))+bcolor(blue)((c)^2....to["using " (2) ]#

#LHS=bc{color(blue)(b+c)}#

#LHS=bc{color(blue)(2a)}...............to["using " (1)]#

#LHS=2abc#

#LHS=RHS#