Taking a,b,c as possible integers, verify the following whether they are identities or not? (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
Answers
Taking a,b,c as possible integers, verify the following whether they are identities or not? (a+b+c)² = a²+b²+c²+2ab+2bc+2ca
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Correction : a² + b² + c² - ab - bc - ca.
= ( a² + b² + c² - ab - bc - ca )
By multiplying it by 2 and dividing it by 2.
= 2 ( a² + b² + c² - ab - bc - ca ) ÷ 2
= ( 2a² + 2b² + 2c² - 2ab - 2bc - 2ca ) ÷ 2
= ( a² + a² + b² + b² + c² + c² - 2ab - 2bc - 2ca ) ÷ 2
= ( a² + b² - 2ab + b² + c² - 2bc + a² + c² - 2ca ) ÷ 2
= [ ( a - b )² + ( b - c )² + ( a - c )² ] ÷ 2
Now , whatever is the value of ( a - b ) , ( b - c ) and ( a - c ) but its square will be always positive, and 2 is also a positive number.
So, the sum of ( a - b )², ( b - c )² and ( a - c )² will be a positive number and if a positive number is divided by a positive number then the result is also a positive number.
So, it is a positive number,hence it can't be a negative number.
Proved.
Hope it helps !
Answer:
answer is [(a-b)^2+(b-c)^2+(a-c)^2]÷2