2^(2x+1)-2^(x-y+3)-2^(x+y+1)+2^3=0 find the x value

Here is the choices
x,y are different from 0.

A)1 B)2 C)3 D)4 E)5

Update: This has been fixed (so x = y = 1).
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Note that we can rewrite this as
2^((x-y)+(x+y)+1) - 2^(x-y+3) - 2^(x+y+1) + 2^3 = 0
==> 2 * 2^((x-y) + (x+y)) - 2^3 * 2^(x-y) - 2 * 2^(x+y) + 2^3 = 0
==> 2 * 2^((x-y) + (x+y)) - 8 * 2^(x-y) - 2 * 2^(x+y) + 8 = 0

So, the clarify matters, let u = 2^(x-y), v = 2^(x+y).

Thus, the equation reduces to 2uv - 8u - 2v + 8 = 0.

This we can factor by grouping:
2uv - 2v - 8u + 8 = 0
==> 2v(u - 1) - 8(u - 1) = 0
==> (2v - 8)(u - 1) = 0
==> v = 4 or u = 1.

Therefore, 2^(x+y) = 4, and 2^(x-y) = 1
==> 2^(x+y) = 2^2, and 2^(x-y) = 2^0
==> x + y = 2, and x - y = 0.

Solving for x and y yields x = 1 and y = 1.
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I hope this helps!

OK...kb did an outstanding job but made a simple math error near the end.
v = 2^(x+y) = 4 = 2^2 so x+y = 2
u = 2^(x-y) = 1 = 2^0 so x-y = 0 or x = y and they both = 1.

BTW...these x and y values check in the equation.

i'll just inssert values for x.