ans this question of 9th standard
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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
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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 !
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