$(r+1)^{\frac{1}{3}}$ transcendental and $s^2-s$ algebraic on $\mathbb{Q}$ then $r$ is transcendental and $s$ algebraic on $\mathbb{Q}$

Question

Let $r \in \mathbb{R}^+ $. $(1)$ Show that $r$ is transcendental on $\mathbb{Q} \iff (r+1)^{\frac{1}{3}}$ is transcendental on $\mathbb{Q}$, and that $(2)$ $r$ is algebraic on $\mathbb{Q} \iff r^2-r$ is algebraic on $\mathbb{Q}$.

$(1) \Rightarrow $ Let $r$ be transcendental on $\mathbb{Q}$ and let $\mathbb{F} = \mathbb{Q}[(r+1)^{\frac{1}{3}}]$, suppose $[\mathbb{F}:\mathbb{Q}]$ to be a finite integer, then $(r+1)^{\frac{1}{3}} \in \mathbb{F}$, hence $r+1=((r+1)^{\frac{1}{3}})^3 \in \mathbb{F}$ and also $r = (r+1)-1 \in \mathbb{F}$: a contraddiction, because it follows that $\mathbb{Q}[r] \subseteq \mathbb{F}$ and $[\mathbb{F}:\mathbb{Q}]=[\mathbb{F}:\mathbb{Q}[r]][\mathbb{Q}[r]:\mathbb{Q}]= \infty$.

$(2) \Rightarrow$ with the same argument of $(1)$ we prove that $r^2-r \in \mathbb{Q}[r]$, so if $r$ is agebraic on $\mathbb{Q}$ then $r^2-r$ is algebraic on it.

I am struggling a little with $\Leftarrow$ implications, any hint would be appreciated, thank you

Answer 1

$\mathbb Q\left((r+1)^\frac13\right)$ is a finite extension of $\mathbb Q(r)$ hence either both are finite extensions of $\mathbb Q$, or none is.

Similarly, $\mathbb Q(s)$ is a finite extension of $\mathbb Q(s^2-s)$ hence...