(a+b)³+(b+c)³+(c+a)³-3(a+b)(b+c)(c+a)
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Sol : ( a + b )³ + ( b + c )³ + ( c + a )³ - 3 ( a + b ) ( b + c ) ( c + a )
Let : ( a + b ) = A , ( b + c ) = B , ( c + a ) = C.
= A³ + B³ + C³ - 3 ABC
Using identity,
x³ + y³ + z³ - 3 xyz = ( x + y + z ) ( x² + y² + z² - xy - yz - za )
= ( A + B + C ) ( A² + B² + C² - AB - BC - CA )
By substituting the values of A , B and C.
= { a + b + b + c + c + a } { ( a + b )² + ( b + c )² + ( c + a )² - ( a + b ) ( b + c ) - ( b + c ) ( c + a ) - ( c + a ) ( a + b ) }
= ( 2 a + 2 b + 2 c ) ( a² + b² + 2 ab + b² + c² + 2 bc + c² + a² + 2 ac - ab - ac - b² - bc - bc - ab - c² - ac - ac - bc - a² - ab )
= 2 ( a + b + c ) ( 2 a² + 2 b² + 2 c² + 2 ab + 2 bc + 2 ca - a² - b² - c² - 3 ab - 3 bc - 3 ac )
= 2 ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
☺☺☺
Sol : ( a + b )³ + ( b + c )³ + ( c + a )³ - 3 ( a + b ) ( b + c ) ( c + a )
Let : ( a + b ) = A , ( b + c ) = B , ( c + a ) = C.
= A³ + B³ + C³ - 3 ABC
Using identity,
x³ + y³ + z³ - 3 xyz = ( x + y + z ) ( x² + y² + z² - xy - yz - za )
= ( A + B + C ) ( A² + B² + C² - AB - BC - CA )
By substituting the values of A , B and C.
= { a + b + b + c + c + a } { ( a + b )² + ( b + c )² + ( c + a )² - ( a + b ) ( b + c ) - ( b + c ) ( c + a ) - ( c + a ) ( a + b ) }
= ( 2 a + 2 b + 2 c ) ( a² + b² + 2 ab + b² + c² + 2 bc + c² + a² + 2 ac - ab - ac - b² - bc - bc - ab - c² - ac - ac - bc - a² - ab )
= 2 ( a + b + c ) ( 2 a² + 2 b² + 2 c² + 2 ab + 2 bc + 2 ca - a² - b² - c² - 3 ab - 3 bc - 3 ac )
= 2 ( a + b + c ) ( a² + b² + c² - ab - bc - ca )
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Anonymous:
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