Splitting field of f(r)=r^3+r^2-2r-1 over Q

\mathbf{Problem:} Let E=\mathbb{Q}(r) where r^3+r^2-2r-1=0. Verify that r'=r^2-2 is also a root of x^3+x^2-2x-1=0. Determined \text{Gal }E/\mathbb{Q}. Show that E is normal over \mathbb{Q}.
\mathbf{Proof:} r^3+r^2-2r-1=0 means (r+1)(r^2-2)=-1. So we only need to prove \displaystyle \frac{-1}{r+1} is another root of f(x)=x^3+x^2-2x-1=0.
Let y=r+1, f(r)=0 means y^3-2y^2-y+1=0.
Let \displaystyle z=\frac{-1}{y}, then \displaystyle\frac {1}{z^3}(z^3+z^2-2z-1)=0, obviously z\neq 0, thus \displaystyle z=\frac{-1}{r+1} is a root of f(x)=0.
f(x) is monic and has no root in \mathbb{Z}, so it is irreducible in \mathbb{Q}. f(x) is minimal polynomial of r. So [E:\mathbb{Q}]=3. \text{Gal } E/\mathbb{Q} must be the cyclic group of order 3. By theorem 4.7 chapter 4 in Jacoboson’s book, E is normal over \mathbb{Q}.
\mathbf{Remark:} Jacobson, Algebra I. p243.

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