A^2 - B^2 ,difference of two squares.?

A^2-B^2

=

[2012^x+2012^(-x)]^2-[2012^x-2012^(-x)]^2

=

[2012^x+2012^(-x)+2012^x-2012^(-x)]

[2012^x+2012^(-x)-2012^x+2012^(-x)]

=

[2(2012^x)][2(2012^-(x)]

=

4[2012^(x-x)]

=

4[2012^0]

=

4

a = 2012^(x) + 2012^(- x)

b = 2012^(x) - 2012^(- x)

= a² - b²

= (a + b).(a - b)

= { [2012^(x) + 2012^(- x)] + [2012^(x) - 2012^(- x)] }.{ [2012^(x) + 2012^(- x)] - [2012^(x) - 2012^(- x)] }

= { 2012^(x) + 2012^(- x) + 2012^(x) - 2012^(- x) }.{ 2012^(x) + 2012^(- x) - 2012^(x) + 2012^(- x) }

= { 2012^(x) + 2012^(x) }.{ 2012^(- x) + 2012^(- x) }

= { 2 * 2012^(x) }.{ 2 * 2012^(- x) }

= 2 * 2012^(x) * 2 * 2012^(- x)

= 4 * 2012^(x) * 2012^(- x) → recall: x^(a) * x^(b) = x^(a + b)

= 4 * 2012^(x - x)

= 4 * 2012^(0) → recall: x^(0) = 1

= 4

If A = 2012^x + 2012^-x and B = 2012^x - 2012^-x

 A^2 - B^2  = 2(2012^x + 2012^-x)

Remember  the difference of two squares. Any two squared numbers with a              negative between them, will factorise. 

NB You cannot do this for a positive value. 

Hence 

A^2 - B^2 factors to (A - B)(A + B) 

Note the signs. 

Hence substituting 

[(2012^x + 2012^-x)-(2012^x - 2012^-x)][  (2012^x + 2012^-x)+(2012^x - 2012^-x)]

Collecting like terms inside each set of square brackets. 

[ 2(2012^-x)][ 2(2012^x] => 

[2 / 2012^x][2(2012^x] => 

Cancel down by '2012^x' 

Hence 

[2][2] = 4 as required. 

What I'd notice is that 2012^-12 = 1/(2012^12).

Using N = 2012^x, you have:

A = N + 1/N

B = N - 1/N

You know from difference-of-squares that A² - B² = (A + B)(A - B), and those two factors

A + B = N + 1/N + N - 1/N = 2N

A - B =  N + 1/N - N + 1/N = 2/N

A² - B² = (A + B)(A - B) = (2N)(2/N) = 4(N/N) = 4

The point of using N instead of 2012^x was to show that the pattern works for *any* nonzero number N and its reciprocal 1/N.  Multiply their sum (N + 1/N) times their difference (N - 1/N) and you'll get 4.

It's been many years. Will show it step by step.

A=2012^x + 2012^-x

B=2012^x-2012^-x

substitute 2012 as N

A=N^x + N^-x

B=N^x - N^-x

Square each of them, add exponents when multiplying same base number, and N^0=1

A^2=N^(2x) + N^-(2x) +2

B^2=N^(2x)+N^-(2x) -2

Subtract one from the other

A^2-B^2= N^(2x) + N^-(2x) +2 - (N^(2x)+N^-(2x) -2)

A^2-B^2= N^(2x) + N^-(2x) +2 - N^(2x)-N^-(2x) +2

A^2-B^2= +2 +2

A^2-B^2 = 4

value=(A-B)(A+B) + 2*2012^(-x)*(2012)^x=4