It can be written as:
\[((1+y)^2-\color{forestgreen}{2x^{2}(1-y^{2})}+x^{4}(1-y)^{2})+\color{forestgreen}{2x^{2}(1-y^{2})}-2x^2(1+y^2)\]
\[=((1+y)-x^2(1-y))^2-4x^2y^2\]
\[=((1+y)-x^2(1-y)+2xy)((1+y)-x^2(1-y)-2xy)\]
\[\small{\color{blue}{Using\space p^2-q^2=(p+q)(p-q)}}\]

Though I managed to factorize the given expression but factored terms are altogether different from that of Rishabh Cool's answer.
The given expression can be written as :-
\[(1+y)^{2}-2x^{2}(1+y^{2})+x^{4}(1-y)^{2}\,=\,( \color{blue}{ 1+y^{2}}+2y)-2x^{2}(\color{blue}{1+y^{2}})+x^{4}( \color{blue}{1+y^{2}}-2y)\]
\[ \implies (\color{blue}{1+y^{2}})+2y-2x^{2}(\color{blue}{1+y^{2}})+x^{4}(\color{blue}{1+y^{2}})-2x^4y \]
\[\implies (\color{blue}{1+y^{2}})(1-2x^{2}+x^{4})+2y(1-x^{4}) \]
\[\implies (\color{blue}{1+y^{2}})(\color{red}{1-x^{2})^{2}}+2y(\color{red}{1-x^{2}})(1+x^{2}) \]
\[\implies (\color{red}{1-x^{2}}) \left( (\color{blue}{1+y^{2}})(\color{red}{1-x^{2}})+2y(1+x^{2}) \right)\]

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## Comments

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TopNewestIt can be written as: \[((1+y)^2-\color{forestgreen}{2x^{2}(1-y^{2})}+x^{4}(1-y)^{2})+\color{forestgreen}{2x^{2}(1-y^{2})}-2x^2(1+y^2)\] \[=((1+y)-x^2(1-y))^2-4x^2y^2\] \[=((1+y)-x^2(1-y)+2xy)((1+y)-x^2(1-y)-2xy)\] \[\small{\color{blue}{Using\space p^2-q^2=(p+q)(p-q)}}\]

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THANK YOU :D

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Though I managed to factorize the given expression but factored terms are altogether different from that of Rishabh Cool's answer. The given expression can be written as :- \[(1+y)^{2}-2x^{2}(1+y^{2})+x^{4}(1-y)^{2}\,=\,( \color{blue}{ 1+y^{2}}+2y)-2x^{2}(\color{blue}{1+y^{2}})+x^{4}( \color{blue}{1+y^{2}}-2y)\] \[ \implies (\color{blue}{1+y^{2}})+2y-2x^{2}(\color{blue}{1+y^{2}})+x^{4}(\color{blue}{1+y^{2}})-2x^4y \] \[\implies (\color{blue}{1+y^{2}})(1-2x^{2}+x^{4})+2y(1-x^{4}) \] \[\implies (\color{blue}{1+y^{2}})(\color{red}{1-x^{2})^{2}}+2y(\color{red}{1-x^{2}})(1+x^{2}) \] \[\implies (\color{red}{1-x^{2}}) \left( (\color{blue}{1+y^{2}})(\color{red}{1-x^{2}})+2y(1+x^{2}) \right)\]

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Nice... :-)

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