You need to firstly show that [tex]R=(\mathbb R,*)[/tex] is a group, which requires that [tex]R[/tex] is closed under [tex]*[/tex], that [tex]*[/tex] is associative,, that [tex]R[/tex] has an identity element, and that every element in [tex]R[/tex] has a corresponding inverse in [tex]R[/tex]. Then for [tex]R[/tex] to be abelian, you also need to show that [tex]*[/tex] is commutative.
But suppose [tex]a>0[/tex] and [tex]b<0[/tex], where [tex]|a|<|b|[/tex]. For instance, take [tex]a=3[/tex] and [tex]b=-4[/tex]. Then
[tex]3*(-4)=\sqrt{3^3+(-4)^3}=\sqrt{27-64}=\sqrt{-37}\not\in\mathbb R[/tex]
which means [tex]R[/tex] is not closed under [tex]*[/tex] and is therefore not a group, and certainly not an abelian one.
