If $$f\left( x \right) = \frac{{{2^x} + {2^{ - x}}}}{2},$$    then $$f\left( {x + y} \right).f\left( {x - y} \right)$$     is equal to :

Solution :
$$\eqalign{ & f\left( {x + y} \right).f\left( {x - y} \right) \cr & = \frac{{{2^{x + y}} + {2^{ - x - y}}}}{2}.\frac{{{2^{x - y}} + {2^{ - x + y}}}}{2} \cr & = \frac{{{2^{2x}} + {2^{2y}} + {2^{ - 2x}} + {2^{ - 2y}}}}{{2 \times 2}} \cr & = \frac{1}{2}\left[ {\frac{{{2^{2x}} + {2^{ - 2x}}}}{2} + \frac{{{2^{2y}} + {2^{ - 2y}}}}{2}} \right] \cr & = \frac{1}{2}\left[ {f\left( {2x} \right) + f\left( {2y} \right)} \right] \cr} $$