How can I find the inverse of this problem u⊞v:=uv−2(u+v)+6 Let V=(2,∞). For u,v∈V and a∈R define vector addition by u⊞v:=uv−2(u+v)+6 and scalar multiplication by a⊡u:=(u−2)^a+2. It can be shown that (V,⊞,⊡) is a vector space over the scalar field R. Find the following:
the sum:
6⊞10=34
the scalar multiple:
−3⊡6=2.015625
the additive inverse of 6:
⊟6=
the zero vector:
0V=3
the additive inverse of x:
⊟x=
As you can see I found the sum and the scalar multiple as well as the zero vector.However, I tried doing the additive inverse of 6 and the additif inverse of x, but I have no idea on how to find the answers.

Before you can find an additive inverse you must find the additive identity.  That is $e\square x = x$ for all $x$.
so $ex -2(e+x) +6 = x$. Solver for $e$
so $e(x-2)=3x -6$ so $e=\frac {3x-6}{x-2} = 3$
The additive identity is $3$.
Verify:  $k\square 3= 3k -2(k+3) + 6 = k -6 + 6 = k$ so that is indeed the indentity.
Now to find the additive inverse of $6$ we must solve:  $x\square 6 = 3$.
So $6x - 2(x+6) + 6 = 3$
$4x =9$
$x = \frac 94$.
SO $x=\frac 94$ is the additive inverse of $6$.
Verify:  $\frac 94 \square 6 = \frac 94\cdot 6 -2(6 +\frac 94) +6= \frac {27}2 - 12-\frac 93 + 6 = 9-12 + 6 = 3$.  It checks.
