Question

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

Answer 1

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 (Proved)

Answer 2

Answer:

yes it is

Step-by-step explanation:

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 (Proved)