Question

Show that :$(9p-5q)^{2} + 180pq = (9p+5q)^{2}$

Answer 1

Answer:

   (9p−5q)2+180pq=(9p+5q)2

LHS=(9p)2−2(9p)(5q)+(5q)2+180pq

       =81p2−90pq+25q2+180pq

       =81p2+90pq+25q2

RHS=(9p+5q)2

       =(9p)2+2(9p)(5q)+(5q)2

       =81p2+90pq+25q2

LHS=RHS

Answer 2

Step-by-step explanation:

$\tt\red {Given :(9p-5q)^{2} + 180pq = (9p+5q)^{2}}$

LHS=$\tt\pink{(9p-5q)^{2} + 180pq}$

$\tt \blue{ \implies81p^{2}-90pq+25q^{2}+180pq \quad \quad \: \bigg[ \dashrightarrow \: Using \: the \: binomial \: theorem \bigg]}$

$\tt \orange{ \implies81p^{2}+90pq+25q^{2} }$

$\tt \purple{ \implies\left(9p+5q\right)^{2} = RHS}$