You should notice that 8 may be expressed as a power of 2, such that:

`8 = 2^3 => 8^x = (2^3)^x => 8^x = (2^x)^3`

You should come up with the following substitution, such that:

`2^x = y => (2^x)^3 = y^3`

Replacing the variable in equation, yields:

`y...

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You should notice that 8 may be expressed as a power of 2, such that:

`8 = 2^3 => 8^x = (2^3)^x => 8^x = (2^x)^3`

You should come up with the following substitution, such that:

`2^x = y => (2^x)^3 = y^3`

Replacing the variable in equation, yields:

`y + y^3 = 10`

You need to move the terms to one side, such that:

`y^3 + y - 10 = 0`

You may split the term 10 such that:

`y^3 + y - 8 - 2 = 0`

You may form the following two groups, such that:

`(y^3 - 8) + (y - 2) = 0`

You should convert the difference of cubes into a product, such that:

`y^3 - 8 = (y - 2)(y^2 + 2y + 4)`

`(y - 2)(y^2 + 2y + 4) + (y - 2) = 0`

Factoring out `y - 2` yields:

`(y - 2)(y^2 + 2y + 4 + 1) = 0`

`(y - 2)(y^2 + 2y + 5) = 0`

Using the zero product rule, yields:

`y - 2 = 0 => y = 2`

`y^2 + 2y + 5 != 0` because `Delta = 4 - 20 < 0`

You need to solve for `x` the equation `2^x = y,` such that:

`2^x = 2`

Equating the exponents, yields:

`x = 1 `

**Hence, evaluating the solution to the given equation, yields **`x = 1.`