$3\mathrm{.}5$ If $(G,**)$ is a group and $x,y,$zinG, then we can unambiguously write $x**y**z$ to

denote either $(x**y)**z$ or $x**(y**z),$ since, by associativity, these are the same

element. Show that

$(x**y**z{)}^{-1}=z^{-1}**y^{-1}**x^{-1}$