Find the value of `int_(0)^(1)root(3)(2x(3)-3x^(2)-x+1)dx`.

Explanation

To solve the integral \( I = \int_{0}^{1} \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx \), we can use a symmetry property of definite integrals. Let's go through the steps:

### Step 1: Define the Integral
Let
\[ I = \int_{0}^{1} \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx \]

### Step 2: Change of Variable
Now, we will use the substitution \( x = 1 - t \). Therefore, \( dx = -dt \). The limits change as follows:
- When \( x = 0 \), \( t = 1 \)
- When \( x = 1 \), \( t = 0 \)

Thus, the integral becomes:
\[ I = \int_{1}^{0} \sqrt[3]{2(1-t)^3 - 3(1-t)^2 - (1-t) + 1} (-dt) = \int_{0}^{1} \sqrt[3]{2(1-t)^3 - 3(1-t)^2 - (1-t) + 1} \, dt \]

### Step 3: Simplify the Expression
Now we simplify the expression inside the cube root:
\[ 2(1-t)^3 = 2(1 - 3t + 3t^2 - t^3) = 2 - 6t + 6t^2 - 2t^3 \]
\[ -3(1-t)^2 = -3(1 - 2t + t^2) = -3 + 6t - 3t^2 \]
\[ -(1-t) = -1 + t \]
\[ +1 = +1 \]

Combining these:
\[ 2 - 6t + 6t^2 - 2t^3 - 3 + 6t - 3t^2 - 1 + t + 1 = -2t^3 + 3t^2 + t - 1 \]

Thus, we have:
\[ I = \int_{0}^{1} \sqrt[3]{-2t^3 + 3t^2 + t - 1} \, dt \]

### Step 4: Combine the Two Integrals
Now we can express \( I \) as:
\[ I = \int_{0}^{1} \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx + \int_{0}^{1} \sqrt[3]{-2x^3 + 3x^2 + x - 1} \, dx \]

This gives:
\[ 2I = \int_{0}^{1} \left( \sqrt[3]{2x^3 - 3x^2 - x + 1} + \sqrt[3]{-2x^3 + 3x^2 + x - 1} \right) \, dx \]

### Step 5: Evaluate the Integral
Notice that the two terms inside the integral are negatives of each other, hence:
\[ \sqrt[3]{2x^3 - 3x^2 - x + 1} + \sqrt[3]{-2x^3 + 3x^2 + x - 1} = 0 \]

Thus:
\[ 2I = \int_{0}^{1} 0 \, dx = 0 \]

### Step 6: Conclusion
Therefore, we conclude:
\[ I = 0 \]

### Final Answer
The value of the integral is:
\[ \int_{0}^{1} \sqrt[3]{2x^3 - 3x^2 - x + 1} \, dx = 0 \]