Mathematics - Sets and Relations Question with Solution | TestHub

$x^3-3x^{2}y-x{y}^2+3y^3=0$

$\Rightarrow x\left(x^2-y^2\right)-3y\left(x^2-y^2\right)=0$

$\Rightarrow \left(x-3y\right)\left(x-y\right)\left(x+y\right)=0$

Now, $x=y \forall \left(x,y\right)\in N\times N$ so its a reflexive relation.
But not symmetric & transitive

Since, $\left(3,1\right)$ satisfies but $\left(1,3\right)$ does not.

Also $\left(3,1\right)\&\left(1,-1\right)$ satisfies but $\left(3,-1\right)$ does not.