Question

ans this question of 9th standard ​

Answer 1

GIVEN, A+B+C=0

A2/BC+B2CA+C2/AB = A3+B3+C3/ABC

WE , KNOW THAT

A3+B3+C3 = ( A+B+C ) ( A2+B2+C2-AB-BC-CA) + 3ABC

IF A+B+C = 0 THEN A3+B3+C3 = 3ABC

3ABC/ABC

3 (ANSWER)

PLZ MARK ME AS BRAINLEST AND FOLLOW ME

Answer 2

Given, a+b+c = 0 => a+b = - c

On cubing both sides, we get : (a+b)³ = (-c)³

Again, (a+b)³ = a³+b³+3ab(a+b)

So, a³+b³+3ab(a+b) = -c³

=> a³+b³-3abc = -c³ [because, a+b = -c]

=> a³+b³+c³ = 3 abc

Using this relation, a³+b³+c³ = 3abc when a+b+c = 0, we will solve the given problem. The solution is given in the picture attached with this answer. Please see it !