Question

if (a^1/3)+(b^1/3)+(c^1/3)=0.then show that (a+b+c)^3=27 abc​

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

Answer 2

(a^1/3)+(b^1/3)+(c^1/3)=0

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

=>(a+b+c)/3=0

=>a+b+c= 0

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

=> a+b+c =0

THE QUESTION IS INAPPROPRIATE.