Question

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

Answer 1

First solve LHS

= 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca

= according to the identity

(a + b + c) =a^2 b^2 c^2 2ab 2bc 2ca

therefore....

(√2a - √2b - √2c)^2

which is not equal to RHS.

IF THAT NOT MAKE SATISFACTORY....

SO.,.

So the question you have entered is not complete. if you please enter your right problem I may help you....

Answer 2

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 )

We know,

( m - n )² = m² + n² - 2mn

(a - b )² + ( b - c)² + ( c - a )²

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