Question

(x+y)whole square × (x-y)whole square

Answer 1

$\underline{\large\bf{\mathfrak{Hello!}}}$

$= > {(x + y)}^{2} \times {(x - y)}^{2} \\ \\ = > ( {x}^{2} + {y}^{2} + 2xy)( {x}^{2} + {y}^{2} - 2xy) \\ \\ = > {x}^{4} + {x}^{2} {y}^{2} - 2 {x}^{3} y + {y}^{4} + {x}^{2} {y}^{2} - 2x {y}^{3} + 2 {x}^{3}y + 2x {y}^{3} - 4 {x}^{2} {y}^{2} \\ \\ = > {x}^{4} + {y}^{4} + 2 {x}^{2} {y}^{2} - 4 {x}^{2} {y}^{2} \\ \\ = > {x}^{4} + {y}^{4} - 2 {x}^{2} {y}^{2} \\ \\ = > {( {x}^{2} - {y}^{2} ) }^{2} \\ \\ = > {[(x - y)(x + y)]}^{2}$

$\boxed{{[(x - y)(x + y)]}^{2} }$

$<marquee> Hope this helps...:)$

Answer 2

$<b> Answer :$

$= > {(x + y)}^{2} \times {(x - y)}^{2}$

$= > ( {x}^{2} + {y}^{2} + 2xy)( {x}^{2} + {y}^{2} - 2xy)$

$= > {x}^{4} + {x}^{2} {y}^{2} + {2x}^{3}y + {y}^{4} + {x}^{2} {y}^{2} - {2xy}^{3} + {2x}^{3}y + {2xy}^{3} - {4x}^{2} {y}^{2}$

$= > {x}^{4} + {y}^{4} - {2x}^{2} {y}^{2} - {4x}^{2} {y}^{2}$

$= > {x}^{4} + {y}^{4} - {2x}^{2} {y}^{2}$

$= > {( {x}^{2} - {y}^{2} ) }^{2}$

$= > {[(x - y)(x + y)]}^{2}$