Question

Hey guys!!

I need some help from u all...

Plz prove the following!!

Answer 1

67.

Given a + b + c = 0

= > a + b = -c

= > b + c = -a

= > c + a = -b  ----- (1)

Now,

$= \ \textgreater \ \frac{(b + c)^2}{3bc} + \frac{(c + a)^2}{3ca} + \frac{(a + b)^2}{3ab}$

$= \ \textgreater \ \frac{a^2}{3bc} + \frac{b^2}{3ca} + \frac{c^2}{3ab}$

$= \ \textgreater \ \frac{a^3 + b^3 + c^3}{3abc}$

$= \ \textgreater \ \frac{3abc}{3abc}$

= > 1.


Hope this helps!