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Find the value

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 $p^{2}-q^{2}=(p+q)(p-q)$

$=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$

$\therefore x^{4}+x^{2} y^{2}+y^{4}=\left(x^{2}+y^{2}+x y\right)\left(x^{2}+y^{2}-x y\right)$

 

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