if a is a arithmetic mean and b, c be the geometric mean between any two positive numbers, then (b^3+c^3)/abc equals....?

• ans: 2
  
  if x,y be two no.s then
  
  x,a,y are in AP => 2a=(x+y)
  
  x,b,c,y are in GP => b=xr , c = xr^2 where r^3 = y/x and r is cr.
  
  (b^3+c^3)= x^3*r^3 + x^3*r^6 = x^3*r^3(1+r^3) = xy(x+y)
  
  bc = xr * xr^2 = x^2 *r^3 = xy
  
  (b^3+c^3)/abc = xy(x+y) / (x+y)/2 * xy = 2
• 10 years agoHelpfull: Yes(35) No(0)
• ie taking x and y are two numbers
  so x,a,y are in ap and AM => 2a=x+y
  and x,b,c,y are in gp
  
  and by taking two possibilities x,b,c are in gp and
  b,c,y are in gp
  so GM for above two is
  b^2=xc and c^2= by
  
  but we need to find
  
  (b^3+c^3)/abc dividing both terms by abc
  
  (b^2/ac)+(c^2/ab)
  we know b^2 and c^2
  
  xc/ac + by/ab
  
  ie x+y/a
  
  subtituting a=x+y/2
  
  so we get 2
  ans ==2
• 10 years agoHelpfull: Yes(3) No(0)
• how can there be 2 geometric mean values???
• 10 years agoHelpfull: Yes(1) No(2)
• i think a os A.M b is G.M c is H.M so a.c=b*b so ans is 1+(c/b)pow3
  a=(x+y)/2 b=sqrt(xy) c is 2xy/x+y
• 10 years agoHelpfull: Yes(0) No(1)
• rakesh is right two g.m possible b=sqrt(xc) c=sqrt(by)
• 10 years agoHelpfull: Yes(0) No(0)