Question

16) If a+b+c= 0 prove that :-
a^2/BC+b^2/AB+C^2/CA=3​

Answer 1

Step-by-step explanation:

We have,

$a + b + c = 0$

$\implies \: a + b = - c$

$\implies \: (a + b) ^{3} = ( - c) ^{3}$

$\implies \: a^{3} + {b}^{3} + 3ab(a + b) = - c^{3}$

$\implies \: a^{3} + {b}^{3} + 3ab( - c) = - c^{3}$

$\implies \: a^{3} + {b}^{3} - 3abc = - c^{3}$

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

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

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

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