Question

plaese solve the given problems​

Answer 1

Answer:

if : x + y + z = 0, then : x³ + y³ + z³ = 3xyz ................ (1)

∵ a¹ʹ³ + b¹ʹ³ + c¹ʹ³ = 0

(a¹'³)³ + (b¹ʹ³)³ + (c¹ʹ³)³ = 3(a¹ʹ³) (b¹ʹ³) (c¹ʹ³)

a + b + c = 3 (abc)¹ʹ³

( a + b + c )³ = [ 3 (abc)¹ʹ³ ]³

( a + b + c )³ = 3³ [ (abc)¹ʹ³ ]³

( a + b + c )³ = 27abc

So option b is correct...