Question

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

Answer 1

Given:

$(a+b+c)=0$

$(a+b+c)^{3}=0$

$⇒a^{3}+b^{3}+c^{3}-abc=0$

$⇒a^{3}+b^{3}+c^{3}=3abc$

Divide throughout by abc

$⇒a^2bc+b^{3}ac+c^{2}ab=3$

Hope it helps!

PLEASE MARK ME BRAINLIST ANSWER......

Answer 2

Answer:

Given :

(a + b + C )= 0

(a + b + c) ^3 = 0

=> a^3 + b^3 + c^3 - abc = 0

=> a^3 + b^3 + c^3 = 0

divide both side by abc :we have;

a^2bc + b^2ab + c^2ca = 3

Step-by-step explanation:

Hope it will help you!!!!!!