Question

What is the formula of (a-b-c)2 ?

Answer 1

Hi ,

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We know the algebraic identity,

( x + y + z )²

= x² + y² + z² + 2xy + 2yz + 2zx

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Now ,

( a - b - c )²

= [ a² + ( - b )² + ( -c )² ]²

= a²+(-b)²+(-c)²+2a(-b)+2(-b)(-c)+2(-c)a

= a² + b² + c² - 2ab + 2bc - 2ca

I hope this helps you.

: )

Answer 2

Answer:

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

Step-by-step explanation:

(a - b - c)²

= (a - b - c)(a - b - c)

= a² - ab - ac - ab + b² + bc - ac + bc + c²

= a² + b² + c² - 2ab + 2bc - 2ac

Similarly, the formula of (a ± b ± c)² is in the form of a² + b² + c² ± 2ab ± 2bc ± 2ac. As you can see, a x -b = -ab, -b x -c = bc, and a x -c = -ac. Then just plug it in to the above form to get

a² + b² + c² - 2ab + 2bc - 2ac.

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

That is the faster way.

Hope this helps!