You need to evaluate the solutions to the given equation, hence, you need to start by converting the rational power into a radical, such that:

`(7x+1)^(1/3) = root(3)(7x+1)`

Replacing `root(3)(7x+1)` for `(7x+1)^(1/3)` in equation yields:

`root(3)(7x+1) - x = 1`

You need to isolate the cube root to the left side, such that:

`root(3)(7x+1)= x + 1`

You need to remove the cube root, hence, you need to raise to cube both sides, such that:

`7x + 1 = (x + 1)^3`

Expanding the cube yields:

`7x + 1 = x^3 + 1 + 3x(x + 1)`

Reducing duplicate members yields:

`7x = x^3 + 3x^2 + 3x => x^3 + 3x^2 + 3x - 7x = 0`

`x^3 + 3x^2 - 4x = 0`

Factoring out x yields:

`x(x^2 + 3x - 4) = 0`

Using zero product rule, yields:

`x = 0`

`x^2 + 3x - 4 = 0`

Using quadratic formula yields:

`x_(1,2) = (-3+-sqrt(9 + 16))/2 => x_(1,2) = (-3+-5)/2`

`x_1 = 1; x_2 = -4`

Testing the values in equation yields:

`x_1 = 1 => root(3)(7+1) = 1 + 1 => root(3)(8) = 2 => 2 = 2` valid

`x_1 = -4 => root(3)(-28+1) = -4 + 1 => root(3)(-27) = -3 => -3 = -3` valid

**Hence, evaluating the solutions to the given equation, yields `x_1 = 1; x_2 = -4` .**