Question

2a² + 2b² + 2C² - 2ab - 2bc - 2ca = (a - b)² + (b - c)² + (c - a)²​

Answer 1

Answer:

Step-by-step explanation:

2a² + 2b² + 2C² - 2ab - 2bc - 2ca = (a - b)² + (b - c)² + (c - a)²​

-> 2a² + 2b² + 2C² - 2ab - 2bc - 2ca = (a²-2ab+b²) + (b²-2ab+c²) + (c²-2ab+a²)

Open all the brackets on RHS.

->2a² + 2b² + 2C² - 2ab - 2bc - 2ca = a²- 2ab + b² + b² - 2bc + c² + c² - 2ca + a²

->  2a² + 2b² + 2C² - 2ab - 2bc - 2ca = 2a² + 2b² + 2c² - 2ab - 2bc - 2ca

Transpose 2a² + 2b² + 2c² from LHS to RHS.

->  -2ab - 2bc - 2ca = -2ab - 2bc - 2ca

When you transpose all terms from LHS to RHS and add, you can cancel everything and you will be left with nothing.

Therefore, 2a² + 2b² + 2C² - 2ab - 2bc - 2ca = (a - b)² + (b - c)² + (c - a)²​

Answer 2

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!