Can anyone tell me what are the steps for this ?

$$\\2^{(x-2)}-4^x+8^x=0\\ 2^x*2^{-2}-2^{2x}+2^{3x}=0\\ 2^x*2^{-2}-(2^{x})^2+(2^{x})^3=0\\ 2^x(2^{-2}-2^{x}+(2^{x})^2)=0\\ 2^x \mbox{ cannot equal zero so}\\ 2^{-2}-2^{x}+(2^{x})^2=0\\ $Let $ y=2^x\\ 2^{-2}-y+y^2=0\\ y^2-y+\frac{1}{4}=0\\\\ y=\frac{1\pm\sqrt{1-1}}{2}=\frac{1}{2}\\ $therefore$\\ \frac{1}{2}=2^x\\ 2^{-1}=2^x\\ x=-1$$

For the top equation:  there is: (√2 - 1) / (√2 - 1)   --->  but this equals 1. So any finite exponent of 1 will work.

For the second equation:

2^(x - 2) - 4^x + 8^x  =  0

--->  2^(x - 2) - (2^2)^x + (2^3)^x  =  0

--->  2^(x - 2) - 2^(2x) + 2^(3x)  =  0     --->  x = -1     (I solved it by graphing; but it checks fairly easily.)