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2015-05-03T18:40:44+08:00

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Hi there! I am not so sure if this is the right answer but I will show it:

First we factor out xd from the equation and get:
xd(y+ x^{3} +x y^{2} -y)=0 \\ xd(x^3+xy^2)=0 \\  x^{2} d(x^2+y^2)=0

So our 1st case is that x^2d=0
Case 1.1 x^2=0 \\ x=0
Case 1.2 d=0

Then our second case would be  x^{2} +y^2=0
x^2=-y^2

Remember that all square numbers are greater than or equal to zero therefore y^2 cannot be negative and cannot be positive because that would make  x^{2} negative so y^2=0 therefore:
x^2=-y^2 \\ x=y=0

So our solution set:
x={0}
y={0}
d={0}


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