Question

(b+c)whole square /3bc +(c+a)whole square /3ac+(a+b) whole square/3ab=1
Proof

Answer 1

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

Step-by-step explanation:

answer is above the following is answer