Question

Simplify

$(s^{2}t+t{q}^2{)}^2-(2stq{)}^2$

Answer 1

We have, $(s^{2}t+t{q}^2{)}^2-(2stq{)}^2$

$=(s^{2}t{)}^2+(t{q}^2{)}^2+2\times s^{2}t\times t{q}^2-4s^2{t}^2{q}^2$ [$\because (a+b{)}^2=a^2+b^2+2ab$]

$=s^4{t}^2+t^2{q}^4+2s^2{t}^2{q}^2-4s^2{t}^2{q}^2$

$=s^4{t}^2+t^2{q}^4-2s^2{t}^2{q}^2$

$=(s^{2}t{)}^2+(t{q}^2{)}^2-2\times s^{2}t\times t{q}^2$

$=(s^{2}t-t{q}^2{)}^2$ [$\because (a-b{)}^2=a^2+b^2-2ab$]