Question

Show that:

$(4pq+3q{)}^2-(4pq-3q{)}^2=48p{q}^2$

Answer 1

To prove: $(4pq+3q{)}^2-(4pq-3q{)}^2=48p{q}^2$

Lets take LHS and then equate it to RHS.
LHS $=(4pq+3q{)}^2-(4pq-3q{)}^2$

$=(4pq+3q+4pq-3q)(4pq+3q-(4pq-3q)$ [Since, $a^2-b^2=(a+b)(a-b)$]

$=\left(8pq\right)(4pq+3q-4pq+3q)$

$=\left(8pq\right)\left(6q\right)$

$=48p{q}^2$

$=$ RHS

Therefore, LHS $=$ RHS

Hence, proved.

(OR)

LHS $=(4pq+3q{)}^2-(4pq-3q{)}^2$

$=(4pq{)}^2+2(4pq\left)\right(3q)+(3q{)}^2-\left[\right(4pq{)}^2-2\left(4pq\right)\left(3q\right)+\left(3q{)}^2\right]$

[Since, $(a+b{)}^2=a^2+2ab+b^2,(a-b{)}^2=a^2-2ab+b^2$]

$=16p^2{q}^2+24p{q}^2+9q^2-16p^2{q}^2+24p{q}^2-9q^2$

$=48p{q}^2$

$=$RHS

Therefore, $(4pq+3q{)}^2-(4pq-3q{)}^2=48p{q}^2$