a^1/3+b^1/3+c^1/3=0
Prove that (a+b+c) ^3=27abc
Answers
Answered by
8
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
Hence proved !
Answered by
6
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
Step-by-step explanation:
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