Solve for real $x$ and $y$

Question

$$x^4+2x^3-y=-\frac{1}{4}+\sqrt3$$$$y^4+2y^3-x=-\frac{1}{4}-\sqrt3$$ I know this problem has something which I am missing(no one's making a 16 degree equation). It probably has some common factor or something. Some hints please. Thanks.

Answer 1

Given $$x^4+2x^3-y = -\frac{1}{4}+\sqrt{3}..............(1)$$

and Given $$y^4+2y^3-x = -\frac{1}{4}-\sqrt{3}...............(2)$$

Now Adding These Two equation, We get

$$x^4+2x^3-x+y^4+2y^3-y = -\frac{1}{2}$$

So $$\left(x^2+x-\frac{1}{2}\right)^2+\left(y^2+y-\frac{1}{2}\right)^2 = 0$$

Which is Possible, When $$x^2+x-\frac{1}{2}=0$$ and $$y^2+y-\frac{1}{2} =0$$

Now Solve these two equation , We get $$\left(x,y\right)\in \left\{-\frac{1}{2}+\frac{\sqrt{3}}{2}\;\;,-\frac{1}{2}-\frac{\sqrt{3}}{2}\right\}.$$

Now After verification, We get $$\left(x,y\right) = \left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\;\;,-\frac{1}{2}-\frac{\sqrt{3}}{2}\right)$$