# Is $d(x,y)=|x^3-y^3|$ metric on $\mathbb R$? [duplicate]

What about property $$2$$:

$$d(x,y)=0$$ iff $$x=y$$

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

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

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

If $$x-y=0$$ then $$x=y$$. But if $$x^2+xy+y^2=0$$ then $$x$$ does not equal to $$y$$.

• Are there any real solutions to $x^2+xy+y^2=0$ other than $(0,0)$? – lulu Nov 22 '19 at 16:48
• Also take a look at math.stackexchange.com/questions/3234082/dx-y-fx-fy-on-mathbbr – Tom Nov 22 '19 at 16:50
• A year too late but $x^2 +xy + y^2 = 0\iff x =\frac {y \pm\sqrt{y^2 -4y^2}}2=\frac {y\pm\sqrt{-3y^2}}2$. If we are restricting to reals then $-3y^2 \le 0$ so this is only possible if $y = 0$ and therefore $x=0$. ... and If we are not restriction to reals, then $|x^3-y^3|$ is not a metric on $\mathbb C$. – fleablood Oct 18 '20 at 19:42

You don’t need to factorize that polynomial to see that there aren’t any nontrivial solutions to the equation $$x^3 - y^3 = 0$$ - just ask yourself, “Are there any pairs of distinct numbers $$x, y$$ such that $$x^3 = y^3$$? The answer is no, because cubing is an injective function. Therefore the property is satisfied
Note that $$x^2+xy+y^2= (x+\frac {y}{2} )^2 + \frac {3}{4}y^2$$
Thus $$x^2+xy+y^2=0 \iff x=y=0$$