You should use the following notation such that: 2^x - 1/(2^(x-1) )= y.

Raising to cube yields:

(2^x - 1/(2^(x-1)))^3 = 2^(3x) - 1/(2^(3(x-1))) - 3*2^x*(1/2^(x-1))*(2^x - 1/(2^(x-1)))

Reducing like powers yields:

(2^x - 1/2^(x-1))^3 = 2^(3x) - 1/(2^(3(x-1))) - 6*(2^x - 1/(2^(x-1)))

Plugging y instead of 2^x - 1/(2^(x-1)) in the equation above yields:

y^3 = 2^(3x) - 1/(2^3(x-1)) - 6y => 2^(3x) - 1/(2^3(x-1)) = y^3 + 6y

You need to write the equation using the new variable such that:

y^3 + 6y - 6y = 1

Reducing opposite terms yields:

y^3 = 1 => (y - 1)(y^2 + y + 1) = 0

You should notice that the factor y^2 + y + 1 is positive all the time, hence only y - 1 = 0.

y - 1 = 0 => 2^x - 1/(2^(x-1)) - 1 = 0

2^x - 2/2^x - 1 = 0

You should use the following notation: 2^x = z.

You need to bring all terms to a common denominator such that:

z^2 - 2 - z = 0 => z^2 - z - 2 = 0

z^2 - z - 2 = z^2 - z - 1 - 1 = (z^2 - 1) - (z+1) = 0

Factoring out z + 1 yields:

(z + 1)(z - 1 - 1) = 0 => z + 1 = 0 => z = -1 => 2^x = -1 is not possible

z - 2 = 0 => z = 2 => 2^x = 2

Equating the exponents yields: x = 1

**Hence, the solution to the equation is x = 1.**