The graph of the equation $x+y=x^3+y^3$ is the union of

Question

The graph of the equation $x+y=x^3+y^3$ is the union of

$(A)$line and an ellipse
$(B)$line and a parabola
$(C)$line and hyperbola
$(D)$line and a point

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I tried to factorize the given equation.

$x^3-x+y^3-y=0$

$x(x^2-1)+y(y^2-1)=0$

The answer given is a line and an ellipse but i do not understand how this equation is split into a line and an ellipse equation.

Answer 1

$$(x+y)=x^3+y^3=(x+y)(x^2-xy+y^2)$$

It can be $\;y=-x\;$ , a straight line, or else, after cancelling:

$$x^2-xy+y^2=1\iff x^2-xy+y^2-1=0\iff$$

$$\left(x-\frac y2\right)^2+\frac34y^2=1\;\;(**)$$

Put now:

$$\begin{cases}u:=x-\frac y2\\{}\\v:=y\end{cases}\iff x=u+\frac v2\;,\;\;y=v$$

and you get the ellipse $\;u^2+\frac34v^2=1\;$

Answer 2

$$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$$$\therefore (x+y)(x^2-xy+y^2-1)=0$$Hopefully you can conclude from here?

Answer 3

Note that $$x+y=x^3+y^3=(x+y)(x^2-xy+y^2)$$ Setting $$x=\cos\left(\frac{\pi}{4}\right)X+\sin\left(\frac{\pi}{4}\right)Y=\frac{X+Y}{\sqrt 2}$$$$y=-\sin\left(\frac{\pi}{4}\right)X+\cos\left(\frac{\pi}{4}\right)Y=\frac{-X+Y}{\sqrt 2}$$ gives $$x^2-xy+y^2=1\Rightarrow 3X^2+Y^2=2.$$