Question

prove that 2a^2 +2b^2 +2c^2 -2ab -2bc -2ca = [ (a-b)^2 + (b-c)^2 + (c-a)^2 ]

Answer 1

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Here is your answer

To prove :

2a² + 2b² + 2c²- 2ab - 2bc -2ca = [ (a-b)^2 + (b-c)^2 + (c-a)^2 ]

Proof :

L.H.S = 2a² + 2b² + 2c² - 2ab - 2bc - 2ca

a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca

a² - 2ab + b² + b² - 2bc + c² + c² - 2ca + a²

( a² - 2ab + b² ) +( b² - 2bc + c² )+ ( c² - 2ca + a² )

Using ( a - b )² = a² - 2ab + b² ,
( b - c )² = b² - 2bc + c² ,
( c - a )² = c² - 2ac + a²

L.H.S = ( a - b )² + ( b - c )² + ( c - a )²

L.H.S = R.H.S

( a - b )² + ( b - c )² + ( c - a )² = ( a - b )² + ( b - c)² + ( c - a )²

Hence Verified

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Answer 2

Answer:

L.H.S.=2a

2

+2b

2

+2c

2

−2ab−2bc−2ca

On re-arranging the terms, we get

=(a

2

−2ab+b

2

)+(b

2

−2bc+c

2

)+(c

2

−2ca+a

2

)

=(a−b)

2

+(b−c)

2

+(c−a)

2

=R.H.S.

Hence, 2a

2

+2b

2

+2c

2

−2ab−2bc−2ca=[(a−b)

2

+(b−c)

2

+(c−a)

2

]