The value of $$\int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} dx$$ is equal to:

We need to find $$\int_0^1 (2x^3 - 3x^2 - x + 1)^{1/3} \, dx$$.

First, let us factor the polynomial inside the cube root.

$$2x^3 - 3x^2 - x + 1$$

Testing $$x = 1$$: $$2 - 3 - 1 + 1 = -1 \neq 0$$.

Testing $$x = -1$$: $$-2 - 3 + 1 + 1 = -3 \neq 0$$.

Testing $$x = 1/2$$: $$2(1/8) - 3(1/4) - 1/2 + 1 = 1/4 - 3/4 - 1/2 + 1 = 0$$. So $$(2x - 1)$$ is a factor.

Dividing: $$2x^3 - 3x^2 - x + 1 = (2x - 1)(x^2 - x - 1)$$.

Now consider the substitution $$x \to 1 - x$$. Let $$g(x) = 2x^3 - 3x^2 - x + 1$$.

$$g(1-x) = 2(1-x)^3 - 3(1-x)^2 - (1-x) + 1$$

$$= 2(1 - 3x + 3x^2 - x^3) - 3(1 - 2x + x^2) - 1 + x + 1$$

$$= 2 - 6x + 6x^2 - 2x^3 - 3 + 6x - 3x^2 - 1 + x + 1$$

$$= -2x^3 + 3x^2 + x - 1 = -(2x^3 - 3x^2 - x + 1) = -g(x)$$

So $$g(1 - x) = -g(x)$$.

Let $$I = \int_0^1 (g(x))^{1/3} \, dx$$.

Using the substitution $$x \to 1 - x$$:

$$I = \int_0^1 (g(1-x))^{1/3} \, dx = \int_0^1 (-g(x))^{1/3} \, dx = -\int_0^1 (g(x))^{1/3} \, dx = -I$$

Therefore $$2I = 0$$, which gives $$I = 0$$.

The answer is $$\boxed{0}$$, which corresponds to Option (1).