Question

If a + b + c = 0, then prove that
$(b + c)^{3} \div 3bc + (c + a) ^{2} \div 3ca + (a + b) {}^{2} \div 3ab$
​

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)