Question:
Find the value
$x^{4}+x^{2} y^{2}+y^{4}$
Solution:
Adding $x^{2} y^{2}$ and subtracting $x^{2} y^{2}$ to the given equation
$=x^{4}+x^{2} y^{2}+y^{4}+x^{2} y^{2}-x^{2} y^{2}$
$=x^{4}+2 x^{2} y^{2}+y^{4}-x^{2} y^{2}$
$=\left(x^{2}\right)^{2}+2 x x^{2}+y^{2}+\left(y^{2}\right)^{2}-(x y)^{2}$
Using the identity $(p+q)^{2}=p^{2}+q^{2}+2 p q$
$=\left(x^{2}+y^{2}\right)^{2}-(x y)^{2}$
Using the identity $p^{2}-q^{2}=(p+q)(p-q)$
$=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$
$\therefore x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$