Solve: `[x]^3 - 2[x] +1 = 0`,

Answer

To solve the equation \([x]^3 - 2[x] + 1 = 0\), where \([x]\) denotes the greatest integer function (also known as the floor function), we can follow these steps:

### Step 1: Substitute \([x]\) with a variable
Let \( n = [x] \). Then, the equation becomes:
\[ n^3 - 2n + 1 = 0 \]

### Step 2: Factor the cubic equation
We can try to find rational roots using the Rational Root Theorem. Testing \( n = 1 \):
\[ 1^3 - 2(1) + 1 = 1 - 2 + 1 = 0 \]
Thus, \( n = 1 \) is a root.

### Step 3: Perform polynomial long division
Now, we can factor the cubic polynomial using \( n - 1 \):
\[ n^3 - 2n + 1 = (n - 1)(n^2 + n - 1) \]

### Step 4: Solve the quadratic equation
Next, we need to solve the quadratic equation \( n^2 + n - 1 = 0 \) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -1 \):
\[ n = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 5: Determine the values of \( n \)
Calculating the roots:
\[ n_1 = \frac{-1 + \sqrt{5}}{2}, \quad n_2 = \frac{-1 - \sqrt{5}}{2} \]
The value \( n_2 \) is negative, and since \( n = [x] \) must be a non-negative integer, we discard \( n_2 \).

### Step 6: Check the validity of \( n_1 \)
Now, we check \( n_1 = \frac{-1 + \sqrt{5}}{2} \):
Calculating \( \sqrt{5} \) gives approximately \( 2.236 \):
\[ n_1 \approx \frac{-1 + 2.236}{2} \approx \frac{1.236}{2} \approx 0.618 \]
Since \( n_1 \) is not an integer, the only integer solution we have is \( n = 1 \).

### Step 7: Find the range of \( x \)
Since \( n = [x] = 1 \), it follows that:
\[ 1 \leq x < 2 \]

### Final Solution
Thus, the solution to the equation \([x]^3 - 2[x] + 1 = 0\) is:
\[ x \in [1, 2) \]