Question

Evaluate:

$(b-c+d+a)(d+a-b+c)+(c-d+a+b)(b+c+d-a)$

• A. $4(ad+bc)$
• B. $2(ad+bc)$
• C. $3(ad+bc)$
• D. $(ad+bc)$

Answer 1

The correct option is A $4(ad+bc)$

$(b-c+d+a)(d+a-b+c)+(c-d+a+b)(b+c+d-a)$

$=\left[\right(d+a)+(b-c\left)\right]\left[\right(d+a)-(b-c\left)\right]+\left[\right(b+c)+(a-d\left)\right]\left[\right(b+c)-(a-d\left)\right]$

$[\because (a-d{)}^2=(d-a{)}^2]$

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

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

$=4ad+4bc$

$=4(ad+bc)$.