a ki cube + b ki cube + c ki cube - 3 abc = 1\ 2( a+ b+ c)[( a- b) whole square + ( b- c ) whole square + ( c- a ) whole square] plz give ans. if u know dont waste my and ur time.
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Heya Dear !!
a³ + b³ + c³ - 3abc = 1/2(a + b + c) [( a - b)² + (b - c)² + (c -a) ²
we know ;
a³ + b³ + c³ = (a + b + c) ( a² + b² + c² - ab - bc - ac )
Multiply and divide by 2 on the RHS
(a + b + c) / 2 ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ac )
(a + b + c) / 2 ( a² + b² - 2ab + b² + c² - 2bc + c² + b² - 2bc )
(a + b + c) / 2 [ (a -b)² + (b - c)² + (c - a)² ]
Hence proved
a³ + b³ + c³ - 3abc = 1/2(a + b + c) [( a - b)² + (b - c)² + (c -a) ²
we know ;
a³ + b³ + c³ = (a + b + c) ( a² + b² + c² - ab - bc - ac )
Multiply and divide by 2 on the RHS
(a + b + c) / 2 ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ac )
(a + b + c) / 2 ( a² + b² - 2ab + b² + c² - 2bc + c² + b² - 2bc )
(a + b + c) / 2 [ (a -b)² + (b - c)² + (c - a)² ]
Hence proved
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