Question

:
lf a +b+c=0, prove that:
a^2/bc+b^2/ab+c^2/ca= 3
​

Answer 1

Answer:

by definition of symmetric function, the correct question is a + b + c = 0 => a²/bc + b²/ca + c²/ab.

(a³ + b³ + c³)/abc ={(a + b + c)(a² + b² + c² - ab - bc - ca) + 3abc/abc = 3abc/abc = 3

Answer 2

Answer:

a²/bc+b²/ac+c²/ab=3 (proved)

Step-by-step explanation:

we know

(a+b+c)³=a³+b³+c³+3(a+b+c)(ab+bc+ca)−3abc

0=a³+b³+c³+0−3abc given ( a +b+c=0)

a³+b³+c³=3abc

a³+b³+c³=3

a²/bc+b²/ac+c²/ab=3

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