[Math] Proving either $x^2$ or $x^3$ is irrational if $x$ is irrational

Question

I had a test today in discrete mathematics and I am dubious whether or not my proof is correct.

Suppose $x$ is an irrational number. Prove that either $x^2$ or $x^3$ is irrational.

My Answer:

Suppose $x^2$ is irrational. If $x^2$ is irrational, then the above statement holds true.
Now suppose $x^2$ is rational. If $x^2$ is rational and since $x$ is irrational, then according to the theorem that a rational number multiplied by an irrational number is irrational,
$x^3=x(x^2)$ is irrational since $x$ is irrational and $x^2$ is rational.
In both cases, the above statement is true, thus it is proved.

I was running out of time and did this really quick and I am not sure if this is really a correct way of doing a proof.

Answer 1

Observe the negation of $p\vee q$ is $\neg p \wedge \neg q$. Suppose both $x^3,x^2$ are rational. Then $x^3/x^2=x$ is?