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.
 A: Hint: As $x>y$.
$x^3-y^3=(x-y)(x^2+y^2+xy)=(x-y)((x+y/2)^2+3y^2/4)>0$ 
A: So, let's use @JMoravitz comment to proceed in the demonstration. We have that

$a>0 \,\,, b>c \implies ab>ac\tag{1}$
$a<0\,\,,b>c \implies ab < ac\tag{2}$
$a>b\,\,,b>c \implies a>c\tag{3}$
$c< b \,\,, b < a \implies c < a\tag{3}$

The first case is when both are positive ($x> 0$ and $y> 0$) and is exactly as he pointed out. We must have $x > y$ as the hypothesis. So
$$x > y \stackrel{x^2>0 \text{ use }(1)}{\implies} x^2>xy \implies x^3 > x^2y \stackrel{y>0,x> y \text{ use }(1)}{\implies}x^3> xy^2 \implies x^3> y^3$$
Let us see the case when both are negative. Then $x< 0$ and $y< 0$. We get that $xx > 0$ and then we use $(1)$
$$x> y \implies x^3> x^2y$$
But now see that $x > y$ so, because that $x< 0$ we have by $(2)$ that $x^2<xy$. And using $(2)$ with $y< 0$ and $x> y$ we have that $xy < y^2$. Using $(3)$ with these two results
$$x^2< xy\tag{1'}$$
$$xy < y^2\tag{2'}$$
We get then using $(3)$ that $x^2 < y^2$
$$x^2 < xy, xy < y^2 \stackrel{(3)}{\implies}x^2< y^2 \stackrel{y< 0}{\implies} x^2y > y^3$$
But then
$$x^3 > x^2 y\,\, ,\,\,  x^2y > y^3 \implies x^3 > y^3$$
A: 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 
$x^2+xy+y^2$.
