Question

Question 5(iv)
Show that:
$(4pq+3q{)}^2-(4pq-3q{)}^2=48p{q}^2$

Answer 1

L.H.S = $(4pq+3q{)}^2-(4pq-3q{)}^2$
$=(4pq{)}^2+2\times 4pq\times 3q+(3q{)}^2-\left[\right(4pq{)}^2-2\times 4pq\times 3q+\left(3q{)}^2\right]$
$\left[Usingidentities\right(a+b{)}^2=a^2+2ab+b^{2}and(a-b{)}^2=a^2-2ab+b^2]$
$=16p^2{q}^2+24p{q}^2+9q^2-[16p^2{q}^2-24p{q}^2+9q^2]$ $=16p^2{q}^2+24p{q}^2+9q^2-16p^2{q}^2+24p{q}^2-9q^2$
$=48p{q}^2$
=R.H.S