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a³+b³+c³-3abc = ½ (a+b+c) {(a-b) ²+(b-c) ²+(c-a) ²}
Answers
a³ + b³ + c³ - 3abc = ½ ( a + b + c) [( a - b)² + (b-c)² + (c - a)² ]
Taking the RHS :
1/2 ( a + b +c) [ (a - b)² + (b - c)² + (c -a)² ]
Solving by using identity (x-y)² = x² + y² - 2xy
= 1/2 ( a+ b + c) [ a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ac]
= 1/2 ( a + b+ c) [ 2a² + 2b² + 2c² - 2 ab - 2bc - 2ca]
Taking out 2 as common :
= 1/2 ( a + b + c) 2 [ a² + b² + c² - ab - bc - ca]
Now , 2 will be cancelled :
= ( a + b + c) [ a² + b² + c² - ab - bc - ca]
Solving the brackets :
= a [ a² + b² + c² - ab - bc - ca] + b [ a² + b² + c² - ab - bc - ca ] + c [ a² + b² + c² - ab - bc - ca]
= a³ + ab² + ac² - a²b - abc - a²c + ba² + b³ + bc² - ab² - b²c - abc + a²c + b²c + c³ - abc - bc² - c²a
= a³ + b³ + c³ - abc - abc - abc + ac² - c²a - a²b + ba² - a² c + a² c + bc² - bc² + ab² - ab² + b²c - b²c
All the bold terms will be cancelled .
= a³ + b³ + c³ - 3 abc
= LHS
•°• a³ + b³ + c³ - 3abc = ½ ( a + b + c) [( a - b)² + (b-c)² + (c - a)² ]
Hence proved
Your answer
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- 3 a b c = 1/2 ( a + b + c ) [ ( a - b ) + ( b - c )^2 + ( c - a ) ^2 ]
Taking the RHS
1/2 ( a + b + c ) [ ( a - b ) ^2 + ( b - c ) ^2 ( c - a ) ^2
solving by using identity
1/2 ( a + b + c ) [ a^2 + b^2 - 2 a b + b^2 + c^2 - 2 b c + c ^2 + a^2 - 2 a c ]
1/2 ( a + b + c ) [ 2a^2 + 2b^2 + 2 c ^2 - 2 a b - 2 b c - 2 c a ]
Take out 2 as Common
1/2 ( a+ b + c )2 [ a ^3 + b^3 + c^2 - ab - bc - ca ]
Now 2 will be cancelled
( a + b + c ) [ a ^2 +. b ^2 + c ^2 - ab - bc - ca ]
Solving brackets
a[ a^2 + b ^2 + c ^2 - ab - bc - ca ] + b [ a ^2 + b ^2 + c ^2 - ab - bc - ca ] + c [ a^2 + b ^2 + c ^2 - ab - bc - ca ]
a^3 + ab ^2 + ac ^2 - a^2 b - abc-a^2 c + ba ^2 + b^3 + c^3 - abc - bc ^2 - c^2a
a^3 + b ^3 + c ^3 - abc - abc - ABC + ac^2- c^2a - a^2b + ba^2 - a^2c + a^2c + bc^2 + ab^2 -ab^2 + b^2c - b^2c
All the bold term will be canceled
a^3 + b^3 + c^3- 3abc
LHC
therefore
a^2 + b ^3 + c^3 - 3 abc = 1/2 ( a + b + c ) [( a - b )^2 + ( b - c )^2 + ( c- a )^2 ]
Hence proved !!