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If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.

If a^3+b^3+c^3=3abc and a+b+c=0, prove that (b+c)^2/3bc+(c+a)^2/3ac+(a+b)^2/3ab=1.
 

Grade:9

1 Answers

Arnab
26 Points
3 years ago
a + b + c = 0     (given)
or,  b + c = - a
or, c + a = - b
or,  a + b = -c
      (b+c)²/3bc + (c+a)²/3ac + (a+b)²/3ab
   = (-a)²/3bc + (-b)²/3ac + (-c)²/3ab
   = (a)²/3bc + (b)²/3ac + (c)²/3ab
   = a³+b³+c³/3abc
   = 3abc/3abc          (\because a³+b³+c³ = 3abc)
   = 1
hence proved.

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