If the inequality 
\[
((x+y)^2+4)((x+y)^2-2)\geq A\cdot (x-y)^2
\]
is hold for every real numbers $x,y$ such that $xy=1$, what is the largest value of $A$?

$ 
\textbf{(A)}\ 12
\qquad\textbf{(B)}\ 14
\qquad\textbf{(C)}\ 16
\qquad\textbf{(D)}\ 18
\qquad\textbf{(E)}\ 20
$
1. **Substitute \( x^2 + y^2 \) in terms of \( t \):**
   Given \( xy = 1 \), we can express \( x^2 + y^2 \) as \( txy \) where \( t \geq 2 \). Therefore, \( x^2 + y^2 = t \).

2. **Rewrite the inequality:**
   \[
   ((x+y)^2 + 4)((x+y)^2 - 2) \geq A \cdot (x-y)^2
   \]
   Using \( x^2 + y^2 = t \) and \( xy = 1 \), we have:
   \[
   (x+y)^2 = x^2 + y^2 + 2xy = t + 2
   \]
   and
   \[
   (x-y)^2 = x^2 + y^2 - 2xy = t - 2
   \]

3. **Substitute these expressions into the inequality:**
   \[
   ((t+2) + 4)((t+2) - 2) \geq A \cdot (t-2)
   \]
   Simplify the left-hand side:
   \[
   (t+6)(t) \geq A(t-2)
   \]

4. **Expand and simplify the inequality:**
   \[
   t^2 + 6t \geq A(t-2)
   \]
   \[
   t^2 + 6t \geq At - 2A
   \]
   \[
   t^2 + 6t - At + 2A \geq 0
   \]
   \[
   t^2 + (6-A)t + 2A \geq 0
   \]

5. **Analyze the quadratic inequality:**
   For the quadratic \( t^2 + (6-A)t + 2A \geq 0 \) to hold for all \( t \geq 2 \), its discriminant must be non-positive:
   \[
   \Delta = (6-A)^2 - 8A \leq 0
   \]

6. **Solve the discriminant inequality:**
   \[
   (6-A)^2 - 8A \leq 0
   \]
   \[
   36 - 12A + A^2 - 8A \leq 0
   \]
   \[
   A^2 - 20A + 36 \leq 0
   \]

7. **Solve the quadratic inequality:**
   \[
   A^2 - 20A + 36 = 0
   \]
   Using the quadratic formula \( A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   A = \frac{20 \pm \sqrt{400 - 144}}{2}
   \]
   \[
   A = \frac{20 \pm \sqrt{256}}{2}
   \]
   \[
   A = \frac{20 \pm 16}{2}
   \]
   \[
   A = 18 \quad \text{or} \quad A = 2
   \]

8. **Determine the largest value of \( A \):**
   Since we are looking for the largest value of \( A \), we choose \( A = 18 \).

The final answer is \( \boxed{18} \)