what is the solution of 4x^4+y^4
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4x^4+y^4
=(2x^2)^2+(y^2)^2 --------------(i)
(x+y)^2=x^2+y^2+2xy
or,x^2+y^2+2xy=(x+y)^2
or,x^2+y^2=(x+y)^2-2xy -------------(ii)
From (i) and (ii) we get,
(2x^2)^2+(y^2)^2=(2x^2+y^2)^2-
4x^2y^2
Hence,4x^4+y^4=(2x^2+y^2)^2-4x^2y^2.
=(2x^2)^2+(y^2)^2 --------------(i)
(x+y)^2=x^2+y^2+2xy
or,x^2+y^2+2xy=(x+y)^2
or,x^2+y^2=(x+y)^2-2xy -------------(ii)
From (i) and (ii) we get,
(2x^2)^2+(y^2)^2=(2x^2+y^2)^2-
4x^2y^2
Hence,4x^4+y^4=(2x^2+y^2)^2-4x^2y^2.
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