If a+b+c = 0 then a^3+b^3+c^3 = 3abc.
Prove this identity.
Answers
Answered by
2
Identity used :
a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
When ( a + b + c ) = 0
a³ + b³ + c³ - 3abc = ( 0 ) ( a² + b² + c² - ab - bc - ca )
a³ + b³ + c³ - 3abc = 0
a³ + b³ + c³ = 3abc
Hence proved!!
a³ + b³ + c³ - 3abc = ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
When ( a + b + c ) = 0
a³ + b³ + c³ - 3abc = ( 0 ) ( a² + b² + c² - ab - bc - ca )
a³ + b³ + c³ - 3abc = 0
a³ + b³ + c³ = 3abc
Hence proved!!
Answered by
1
Answer:
a + b + c = 0
a³ + b³ + c³ - 3abc = (a+b+c)(a² + b² + c² - ab - bc - ca)
= 0(a² + b² + c² - ab - bc - ca)
= 0
a³ + b³ + c³ - 3abc = 0
a³ + b³ + c³ = 3abc
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