can someone help with simplifying this problem?

Question

(3^x+3^-x)^2-(3^x-3^-x)^2
the answer is 4, but i dont know how to get it

Answer 1

a^2 - b^2 = (a-b)(a+b)
(3^x+3^-x)^2-(3^x-3^-x)^2 =
(3^x + 3^-x + 3^x - 3^-x)(3^x + 3^-x - 3^x + 3^-x)
= (2*3^x)(2*3^-x) = 4 * 3^0 = 4

Answer 2

Let m = 3^x.

Now the problem looks like this:

(m + 1/m)^2 - (m - 1/m)^2

which is the difference of two squares.
a^2 - b^2 = (a + b)(a - b) so this factors as

(m + 1/m + m - 1/m) * (m + 1/m - m + 1/m) =
(2m) * (2/m) = 2 * 2 * m * 1/m = 4.

Answer 3

let y= 3^x


therefore ,

(3^x+3^-x)^2-(3^x-3^-x)^2
= ( y + y^-1 )^2 - ( y - y^-1 ) ^2
= ( y+y^-1 + y - y^-1 ) ( y + y^-1 - y + y^-1) [ (a+b)(a-b)=a^2-b^2]
= (2y) ( 2 y^-1)
= 2 * y * 2 * y^-1
= 4 [ y * y^-1 = y * 1/y = 1]

Hence proved.

Answer 4

(3^x+3^-x)^2-(3^x-3^-x)^2
(3^2x +2*3^0 +3^-2x) - (3^2x -2^3^0 +3^-2x)
=3^2x +2 +3^-2x -3^2x+2-3^-2x
= 2+2 = 4