Question

If a, b, c are all non zero and a+b+c =0. prove that a square \bc + b square \ca +c square \ab = 3.

Answer 1

We know that if
$a + b + c = 0$
Then,

$a ^{3} + {b}^{3} + {c}^{3} = 3abc$

Now, dividing both sides by abc, we get

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

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

Proved

That's your answer.

Hope it'll help.. :-)