How a³+b³+c³-3abc = (a+b+c )(a²+b²+c²-ab-bc-ca)
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Posted by Tirthesh Shah 5 years, 2 months ago
- 1 answers
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Sia ? 5 years, 2 months ago
First of all you should be know that
formula
a³ + b³ + c³ -3abc = ( a + b + c)(a² + b² + c² - ab - bc - ca).
LHS = a³ + b³ + c³ - 3abc
= ( a³ + b³ ) + c³ - 3abc
= ( a + b)³ - 3ab( a + b) +c³ - 3abc
= ( a + b)³ - 3a²b - 3ab² + c³ - 3abc
= {(a + b)³ + c³ } -3a²b - 3ab² - 3abc
= ( a + b + c)³ - 3c(a + b) ( a + b + c ) -3ab(a + b+ c)
= ( a + b + c){ ( a + b + c)² -3c(a +b) - 3ab }
= ( a + b + c){ a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab }
= ( a + b + c)( a² + b² + c² - ab - bc -ca) = RHS
LHS = RHS
Hence, Proved
0Thank You