<s>[INST] Proofs Involving Real Numbers We have the following rules that must be used in solving the question

  
*
  
*$a>0$ and $b>c$ $\implies$ $ab>ac$
  
*$a<0$ and $b>c$ $\implies$ $ab<ac$
  
*$a>b$ and $b>c$ $\implies$ $a>c$
  

Prove that the following is true: $\forall x,y \in \mathbb{R}$ is true that 

$$ x>y \implies x^3>y^3$$

Distinguish the cases according to the signs of $x$ and $y$.
I know that there are 4 such cases. 


*

*$+x$ and $+y$

*$+x$ and $-y$

*$-x$ and $+y$

*$-x$ and $-y$


I have already ruled out that (3) is not possible as this would contradict the whole proof. But after this, I do not know where to proceed.
 [/INST] Another way:
$\begin{array}\\
x^3-y^3
&=(x-y)(x^2+xy+y^2)\\
&=\frac12(x-y)(2x^2+2xy+2y^2)\\
&=\frac12(x-y)(x^2+2xy+y^2+x^2+y^2)\\
&=\frac12(x-y)((x+y)^2+x^2+y^2)\\
\end{array}
$
so
$x^3-y^3$
has the same sign as
$x-y$.
I prefer this
rewriting of
$x^2+xy+y^2$
as
$\frac12((x+y)^2+x^2+y^2)$
(rather than
$(x+y/2)^2+3y^2/4$)
because it is
symmetrical in
$x$ and $y$
like 
$x^2+xy+y^2$.
</s>