Question

please solve it is urgent​

Answer 1

Answer:

Step-by-step explanation:

We have,

a + b + c = 0

To prove that, a^2(b+c)+b^2(c+a)+c^2(a+b)+3abc=0

∵ a + b + c = 0

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

L.H.S. = (b + c)+ (c + a) + (a + b) + 3abc

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

= (- a)+ (- b) +$c^{2}$(- c) + 3abc

= -a^{3} -b^{3}-c^{3} + 3abc

= - (a^{3}+b^{3}+c^{3} - 3abc)

We also know that $a^{3}$+$b^{3}$+$c^{3}$- 3abc = (a + b + c)(a^{2} +b^{2} +c^{2} -ab-bc-ca)

= - (a^{3}+b^{3}+c^{3} - 3abc)= (0)(a^{2} +b^{2} +c^{2} -ab-bc-ca)

=0

hence proved

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