Question

plz tell answer..of ​

Answer 1

Answer:

$a + b + c = 0 \\ so \\ a + b = - c \\ b + c = - a \\ c + a = - b$

On putting values

$\frac{ {(b + c)}^{2} }{3bc} + \frac{ {(c + a)}^{2} }{3ca} + \frac{ {(a + b) }^{2} }{3ab} \\ = \frac{ { - a}^{2} }{3bc} + \frac{ { - b}^{2} }{3ca} + \frac{ { - c}^{2} }{3ab} \\ = > taking \: lcm \\ = \frac{ { - a}^{3} - {b}^{3} - { c }^{3} }{3abc} \\ - 1 \: is \: common \: so \\ = \frac{ {a}^{3} + {b}^{3} + {c}^{3} }{ - 3abc}$

As identity says

As identity says if x + y + z = 0

As identity says if x + y + z = 0then x³ + y³ + z³ = 3xy

So a³ + b³ + c³ = 3abc [ As a + b + c = 0 ]

So answer will be

$= \frac{3abc}{ - 3abc} \\ = - 1 \: ans.$

#ßr@!ηl¡e$t

Answer 2

Answer:

-1

Step-by-step explanation:

3abc/-3abc

=-1

the ans is -1