Question

If a+b+c=0 ,then prove that (b+c)²/3bc + (c+a)²/3ac + (a+b)²/3ab =1

Answer 1

(b+c)²/3bc+(c+a)²/3ac+(a+b)²/3ab
=(b²+2bc+c²)/3bc+(c²+2ac+a²)/3ac+(a²+2ab+b²)/3ab
=(ab²+2abc+ac²+bc²+2abc+a²b+a²c+2abc+b²c)/3abc
={ab(a+b)+bc(b+c)+ac(a+c)+6abc}/3abc
=(-abc-abc-abc+6abc)/3abc [∵, a+b+c=0,∴,a+b=-c,b+c=-a,a+c=-b]
=(6abc-3abc)/3abc
=3abc/3abc
=1 (Proved)

Answer 2

Answer:

Step-by-step explanation:(b+c)²/3bc+(c+a)²/3ac+(a+b)²/3ab

=(b²+2bc+c²)/3bc+(c²+2ac+a²)/3ac+(a²+2ab+b²)/3ab

=(ab²+2abc+ac²+bc²+2abc+a²b+a²c+2abc+b²c)/3abc

={ab(a+b)+bc(b+c)+ac(a+c)+6abc}/3abc

=(-abc-abc-abc+6abc)/3abc [∵, a+b+c=0,∴a

+b=-c,b+c=-a,a+c=-b]

=(6abc-3abc)/3abc

=3abc/3abc

=1