Question

If a+b+c = 0, then write the value of
$\frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ab}$
​

Answer 1

Step-by-step explanation:

$\frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ab} \\ = \frac{a {}^{2} \times a }{bc \times a} + \frac{b {}^{2} \times b }{ca \times b} + \frac{c {}^{2} \times c}{ab \times c} \\ = \frac{a {}^{3} }{abc} + \frac{b {}^{3} }{abc} + \frac{c {}^{3} }{abc} \\ = \frac{{a}^{3} + {b}^{3} + {c}^{3} }{abc} \\ = \frac{(a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ac) + 3abc}{abc} \\ = \frac{0 \times ( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ac) + 3abc}{abc} \\ = \frac{0 + 3abc}{abc} \\ = \frac{ 3abc}{abc} = \: 3 \\ thus \\ \frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ca} + \frac{c {}^{2} }{ab} \: = 3$

Mark it as brainliest.

Answer 2

Step-by-step explanation:

You know this is very simple than the logic you have given earlier..... Just see

$we \: know \: if \: (a + b + c) = 0 \\ then \: a {}^{3} + b {}^{3} + c {}^{3} - 3abc = 0 \\ a {}^{3} + b {}^{3} + c {}^{3} = 3abc \\ \frac{a {}^{3} + b {}^{3} + c {}^{3} }{abc} = 3 \\ \frac{a {}^{3} }{abc} + \frac{b {}^{3} }{abc} + \frac{c {}^{3} }{abc} = 3 \\ \frac{a {}^{2} }{bc} + \frac{b {}^{2} }{ac} + \frac{c {}^{2} }{ab} = 3$

That is the value is 3