Question

The simplified value of the experession${(a+b-c)}^2+2(a+b-c)(a-b+c)+{(a-b+c)}^2$ is

• A. $0$
• B. $4a^2$
• C. $4b^2$
• D. $c^2$

Answer 1

Answer: (B)

$4a^2$

Since $(x+y{)}^2=x^2+y^2+2xy$,

In the expression $(a+b-c{)}^2+2(a+b-c\left)\right(a-b+c)+(a-b+c{)}^2$,

$x=a+b-c$ and $y=a-b+c$

Therefore, using the identity $(x+y{)}^2=x^2+y^2+2xy$, the expression $(a+b-c{)}^2+2(a+b-c\left)\right(a-b+c)+(a-b+c{)}^2$ can be written as:

$(a+b-c{)}^2+2(a+b-c\left)\right(a-b+c)+(a-b+c{)}^2=\left[\right(a+b-c)+(a-b+c){]}^2=(2a{)}^2=4a^2$

Hence, $(a+b-c{)}^2+2(a+b-c\left)\right(a-b+c)+(a-b+c{)}^2=4a^2$.