If S is sum of x^3 (x is root of equation 2^(x^2-4x-1)+2^(x^2-4x-2)=24), what is S?

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You need to evaluate the summation of the cubes of solutions to the exponential equation, hence, you need to find the solution to this equation. You should come up with the following substitution, such that:

`2^(x^2-4x) = t`

Using the rules of exponentiation, such that:

`t*2^(-1) + t*2^(-2) = 24`

`t/2 + t/(2^2) = 24 => t/2 + t/4 = 24 `

Bringing the terms to a common denominator, yields:

`2t + t = 4*24 => 3t = 4*24 => t = 4*8 => t = 32`

You need to solve for x the following equation, such that:

`2^(x^2-4x) = t => 2^(x^2-4x) = 32 => 2^(x^2-4x) = 2^5`

Equating the powers yields:

`x^2 - 4x = 5 => x^2 - 4x - 5 = 0`

`x_(1,2) = (4+-sqrt(16 + 20))/2 => x_(1,2) = (4+-6)/2`

`x_1 = 5 ; x_2 = -1`

You need to evaluate the cubes of the roots, such that:

`(x_1)^3 = 5^3 => (x_1)^3 = 125`

`(x_2)^3 = (-1)^3 => (x_2)^3 = -1`

You need to evaluate the summation of the cubes of the roots, such that:

`S = (x_1)^3 + (x_1)^3 => S = 125 - 1 => S = 124`

Hence, evaluating the summation of the cubes of the roots, under the given conditions, yields `S = 124.`