$xI'm looking for a proof to the following theorem:

For any $x,y\in R$: $x<y \Rightarrow x^3<y^3$

I'm trying this approach:

Let $z = x^3 - y^3 = (x-y)(x^2+xy+y^2) = z_1 z_2$

where $z_1 = x-y$ and $z_2 = x^2+xy+y^2$, so I must show that $z<0$. I know that $x<y \iff x-y<0$ so I just have to show that $z_2>0$ for any $x,y \in R$
I find it easy to prove that $z_2 > 0$ in the cases when $0<x<y$ and when $x<y<0$, but I can't prove it when $x<0<y$.
Here's my draft:

$x<y \iff x^2>xy$ $(1)$
$y<x \iff y^2>xy$ $(2)$
$(1)+(2)$ makes $x^2+y^2>2xy \iff x^2+xy+y^2>3xy < 0$

Any hints?
Thank you in advance!
 A: You don't need all that high-powered algebra. (By that I mean the factorization and the cleverness Kaster helped you with.) All you need is that if $a<b$ and $c>0$, then $ac<bc$. It's more boring, but it doesn't require any cleverness.
With that in mind, here's another proof, in three cases.
Case one: $xy>0$
If we multiply $x<y$ by the positive quantities $x^2$, $xy$, and $y^2$, we get the following three inequalities, respectively:
$$x^3<x^2y$$
$$x^2y<xy^2$$
$$xy^2<y^3$$
Putting these together gives the claim.
Case two: $xy<0$
In this case, we must have $x<0$ and $0<y$. Multiplying the first inequality by the positive quantity $x^2$ gives $$x^3<0$$ and similarly, multiplying the second inequality by $y^2$ gives $$0<y^3$$
Putting these together gives the claim.
Case three: $xy=0$
Finally, we have either $x=0$ or $y=0$ (but not both!). I'll let you deal with this case.
A: $$
x^2 + xy + y^2 = x^2 + xy + \frac {y^2}4 + \frac {3y^2}4 = \left(x + \frac y2 \right )^2 + \frac {3y^2}4 > 0
$$
A: If $x < y$ then we can write $y = x + \epsilon, \epsilon > 0$. Now
$$
\begin{align}
y^3 &= (x + \epsilon)^3\\
&= x^3 + 3x^2\epsilon + 3x\epsilon^2 + \epsilon^3
\end{align}
$$
So we just need to show that for $ \epsilon > 0$, $3x^2 + 3x\epsilon + \epsilon^2 >0$. The discriminant of this equation is $9\epsilon^2 - 12\epsilon^2 < 0$, so there are no real roots as the equation is positive all of the time, which proves that given $y > x, y^3 > x^3$.
A: Here's another way to prove that $x^2+xy+y^2>0$
$$x^2+xy+y^2>0\Leftrightarrow (x+y)^2>xy$$
If you think a little about that inequality you'll see it is always true given that $x \neq y$. To give a formal proof some case working is needed but it's bearable. 


*

*If $x$ or $y$ is zero then the inequality holds (RHS is positive, LHS is zero). 

*If $x$ and $y$ have opposite signs then the inequality also holds (RHS non-negative, LHS negative). 


Now we prove the inequality holds if $x$ and $y$ have the same sign.


*

*If both are positive, $|y|>|x|$:


$$(x+y)^2=(|x|+|y|)^2>|y|^2>|x|\cdot|y|=xy$$


*

*If both are negative, $|x|>|y|$:
$$(x+y)^2=(|x|+|y|)^2>|x|^2>|x|\cdot|y|=xy$$


It is not as elegant as Kaster's solution, but it's a systematic way of getting the job done.
