16+(x^3+4) (x^3-4)

6+(x^3+4) (x^3-4)

 

\(6+(x^3+4) (x^3-4)\\ = (x^3+4) (x^3-4) \;\;\;\;+6\)

 

 

See how yow you have the same thing in both brackets but one is plus and one is minus?

This is a difference ot 2 squares.

and

\((a-b)(a+b) =  a^2-b^2\)

 

so

 

\(= (x^3+4) (x^3-4) \;\;\;\;+6\\ = (x^3)^2\;-\;(4)^2 \;\;\;\;+6\\\)

 

Andre you can try finishing it now :)