Image of an exam question I am revising link: [1]
For (i) I have stated the relation is reflexive as $\forall x ∈ \Bbb R, xPx$ is reflexive as $x^3 \ge x $
For (ii) I have stated that the relation is not symmetric as $\forall x,y ∈ \Bbb R, xRy$ does not equal $yRx$, as for the values $x=2, y=1, x^3 - y \ge y^3 - x$ does not equal to $y^3 - x \ge x^3 - y$
For (iii) I have stated that the relation is transitive as as $xPy$ and $yPz$ imply $xPz \forall x,y,z ∈ P,$ as $\forall x,y,z$ that are elements of $R$ and that satisfy $P$, $(a,b)$ $a$ must always be greater than $b$, therefore $\forall x,y ∈ P x\ge y, \forall y,z ∈ P y\ge z,$ therefore $x\ge z$
The last one is a tad all over the place, but I just wanted to know if I am on the right track and if my answers are correct, if not I would love to know why so I can improve myself and see where I went wrong as to not make the same mistake in the future. Any and all help would be greatly appreciated.
Your resolution is unclear and incorrect in some places.
For reflexivity, that's not what you have to prove (which is also wrong! as, if $x=0.5$ then $x^3\not\ge x$), you have to show that $x^3-x\ge x^3 -x$ which is obviously true (as the expressions are equal).
Also, saying "$xRy$ does not equal $yRx$" doesn't make much sense, you should use a word like equivalent to describe that two propositions imply the same things.
Here's perhaps an easier resolution.
Note that
$$xPy\iff x^3-y\ge y^3-x\iff x^3+x\ge y^3+y.$$
From that it's clear that the relation is reflexive (explained above).
Now, Let $x=1,y=0$, we have that $2=x^3+x\ge y^3+y=0$, and so $xPy$, but $yPx$ is not true (why?), thus $P$ is not symmetric (in fact, it's asymmetric!).
Now, for transitivity, suppose $xPy$ and $yPz$, this says that $x^3+x\ge y^3+y$ and $y^3+y\ge z^3+z$, can you follow?