Don't guess! Take the challenge

Statement $1$ is not always true $√A$ is imaginary if $A<0$ . Statement $2$ is always true as the square root of a positive irrational number is always irrational. Statement $3$ is not always true since $x^3$ could be a negative integer; the problem never states that $x>0$ .

Statement 1:

$\sqrt{A}$ is well-defined since $A$ is positive. So either $\sqrt{A}$ is rational or irrational. But if $\sqrt{A} = \frac{p}{q}$ for $p,q$ integers, then $A = \frac{p^2}{q^2}$ . We are told that $A$ is irrational, thus $\sqrt{A}$ must also be irrational.

Statement 2:

If $n = x^3, |x^3| = y^2$ for $x,y$ integers, then there are two possibilities:

a) $n < 0$ , in which case $\sqrt{n}$ is not well-defined

b) $n \geq 0$ , in which case $\sqrt{n}$ is rational.

Without bothering to prove that these are the only two possibilities, I'll note that $n = -64$ shows that this statement need not always be true. In this case, $x = -4, y = \pm 8,$ but $\sqrt{-64}$ is not well-defined.