Question

If a^1/3+b^1/3+c^1/3 = 0 then prove that (a+b+c)^3 =27abc

Answer 1

Answer:

Step-by-step explanation:

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