Question

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

Answer 1

Using the identity $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$.

Given : ${(a-b+c)}^2-{(a-b-c)}^2$ we apply the above identity as follows:

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

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

$=(a^2+b^2+c^2-2ab-2bc+2ca)-(a^2+b^2+c^2-2ab+2bc-2ca)$

$=a^2+b^2+c^2-2ab-2bc+2ca-a^2-b^2-c^2+2ab-2bc+2ca=-4bc+4ca=4(ca-bc)$

Hence, ${(a-b+c)}^2-{(a-b-c)}^2=4(ca-bc)$.