Question

please solve it second number ​

Answer 1

Answer:

a+b+c=0

Therefore (a+b+c)^2=0....because 0^2=0

a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0

Answer 2

Step-by-step explanation:

We have,

a + b + c = 0

Then,

⇒ a + b = - c, b + c = - a and c + a = - b

L.H.S. = a²(b + c) + b²(c + a) + c²(a + b) + 3abc = 0​

Put a + b = - c, b + c = - a and c + a = - b, we get

= a²(-a)+ b²(- b) + c²(- c) + 3abc

=  -a³ - b³ - c³ + 3abc

= -(a³ + b³ + c³ - 3abc)

We know that,

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

⇒ a³ + b³ + c³ - 3abc = 0

⇒ 0

Hence proved!

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