Question

I'd a,b,c are non zero and a+ b+c =0 than prove that asq/cb+ bsq/ca+csq/ab =3​

Answer 1

Answer:

$we \: know \: that \\ {a}^{3} + {b}^{3} + {c}^{3} - 3abc \\ = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) \\ in \: question \: given \: that \\ a + b + c = 0 \\ then \\ {a}^{3} + {b}^{3} + {c}^{3} = 3abc \\ divide \: abc \: to \: both \: side \\ \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{abc} = 3 \\ \frac{ {a}^{2} }{bc} + \frac{ {b}^{2} }{ac} + \frac{ {c}^{2} }{ba} = 3 \\ hence \: proved$