Define X=R{k} and define ⋆ to be the operation such that x⋆y=(x−k)(y−k)+k. Check each of the four axioms of a group (closure, associative, identity, inverse). Which of them hold? Is (X,⋆) a group.
It says that X can't contain the number k, thus X would be a group only if x ⋆ y != k. But theres a k in the definition of the operation. does that mean that it automatically isnt closed? if not, why ? because ive found that x ⋆ y = k if and only if x =k or y = k which is impossible in the first case which means it is closed. also, can an inverse or identity be not in the set by chance ?
As you said the result of $x\star y$ can only be $k$ if $x=k$ and $y=k$, but since that's not allowed you can conclude that $x\star y = k$ is impossible, thus $(X,\star)$ is closed.
To finde the identity element $1_\star$ you have to solve \begin{align*} x \overset{!}{= }x \star 1_\star = (x-k)(1_\star-k)+k \end{align*} for $1_\star$ which should result in $1_\star = k+1$ (which is not $k$).
In order to find inverse elements you have to solve \begin{align*} k+1 = 1_\star \overset{!}{= }x \star x^{-1} = (x-k)(x^{-1}-k)+k \end{align*} for $x^{-1}$ which should result in $x^{-1} = \frac{1}{x-k}+k$ (which can never be $k$)
In order to check for associativity you have to check that \begin{align*} ((x-k)(y-k)+k)(z-k)+k = (x\star y)\star z \overset{!}{=} (x-k)((y-k)(z-k)+k)+k \end{align*} which will turn out to be not true.