Question

formula for [a-b-c] whole square?

Answer 1

To find :

(a-b-c)²

Solution :

(a - b - c )²

This can be split up as,

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

Multiplying both the terms, we get,

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

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

= a² + b² + c² - 2ab + 2bc - 2ac

Therefore,

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

Answer 2

Answer:

Expansion of (a-b-c)²

(a-b-c)²=a²+b²+c²-2ab+2bc-2ca

Step-by-step explanation:

Expansion of (a-b-c)²

=[a+(-b)+(-c)]²

/* By algebraic identity:

$\boxed {(x+y+z)^{2}\\=x^{2}+y^{2}+z^{2}+2xy+2yz+2zx}$*/

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

=a²+b²+c²-2ab+2bc-2ca

Therefore,

Expansion of (a-b-c)²

=a²+b²+c²-2ab+2bc-2ca

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