Question

If a3+b3+c3=3abc & a+b+c=0 show 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

a+b+c=0; so, a+b=-c..........



=(-abc-abc-abc+6abc)/3abc
=(6abc-3abc)/3abc
=3abc/3abc
=1 (Proved)

hope it helps you

Answer 2

Answer:

Step-by-step explanation:

Refer to the image below