How to find the sum of all elements of inverse of a matrix without finding the inverse explicitly? Problem Statement:

Find all real $x$ so that the matrix is invertible. $$\begin{bmatrix}
 2+x & 2 & 2 & 2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x & 2 \\ 2 & 2 & 2 &
 2+x \end{bmatrix}$$
Assuming that $A^{-1}$ exists find the sum of all the elements of
  $A^{-1}$ without explicitly computing $A^{-1}$.

For the first part I calculated the determinant (came out to be $x^4+8x^3$) and put it to $0$ to find that the matrix is not-invertible if $x=-8$ or $x=0$. But I've not idea how to approach the second part of the problem. 

As suggested here, the sum of all elements in a matrix $M$ is given by $\langle Me, e\rangle$ where $e = \pmatrix{1 \\ 1 \\ 1 \\ 1}$.
In that case, if $v$ is a vector such that $Av = e$, then we have $$\langle A^{-1}e, e\rangle = \langle v, e\rangle$$
Therefore, we have to solve the system
$$\pmatrix{2+x & 2 & 2 &2 \\ 2 & 2+x & 2 & 2 \\ 2 & 2 & 2+x &2 \\ 2 & 2 & 2 & 2+x} \pmatrix{v_1 \\ v_2 \\ v_3 \\ v_4} = \pmatrix{1 \\ 1 \\ 1 \\ 1}$$
Since $A$ is invertible, the solution is unique and it is given by $$v = \frac{1}{x+8}\pmatrix{1 \\ 1 \\ 1 \\ 1}$$
Therefore the sum is equal to 
$$\langle v, e\rangle = \frac1{x+8}\|e\|^2 = \frac{4}{x+8}$$
Note that solving the linear system above is in general faster than inverting the matrix, especially in this case since your $A$ is symmetric.
