Question

A plus B minus C whole square

Answer 1

$(a+b-c)^{2} \text { is } \bold{a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a}$

To find:

a plus b minus c whole square = ?

Solution:

Given: $(a+b-c)^{2}$

We know that  

$(a+b+c)^{2}$

$=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a$

This is the final answer to the question (a+b+c)^{2} and a well-established solution or formula. But we are to find out the solution with a slight change in these quadratic polynomial.

$(a+b-c)^{2}$

$=a^{2}+b^{2}+(-c)^{2}+2 a b+2 b(-c)+2(-c) a$

$=a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a$

Hence, the final answer to the question $(a+b-c)^{2} \text { is } a^{2}+b^{2}+c^{2}+2 a b-2 b c-2 c a$

Answer 2

Answer:

(A+B-C)² = A²+B²+C²+2AB-2BC-2CA

Explanation:

We know the algebraic identity:

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

Now ,Here

x = A , y = B, z = -C

(A+B-C)²

[A+B+(-C)]²

= A²+B²+(-C)²+2AB+2B(-C)+2(-C)A

= A²+B²+C²+2AB-2BC-2CA

Therefore,

(A+B-C)²

=A²+B²+C²+2AB-2BC-2CA

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