To do this requires thinking of the fact that  

(A+)² = (A+)(A+) = A² + 2A·( +  = A² + 2 +  = A² +  + 2.

Therefore

A² +  + 2 = (A+)²

or

A² +  = (A+)² - 2
 
The theorem which the above proves:

The sum of a square and its reciprocal equals to the square of the
sum of their two square roots decreased by 2

x4 +  

That's the sum of a square and its reciprocal, so using the 
theorem above it becomes

(x² + )² - 2

What's inside the parentheses is also the sum of a square
and its reciprocal, so using the theorem above it becomes

[(x + )² - 2]² - 2

Now we need to find 

                 by substituting 3 + 2 for x  
                
                 =  =  =
                 =  =  =
                  = 3 - 2

Therefore 

[(x + )² - 2]² - 2

becomes

[(3 + 2 + 3 - 2)² - 2]² - 2

[(6)² - 2]² - 2

[36 - 2]² - 2

34² - 2

1156 - 2

1154

Edwin