To solve the equation \( x(x-1)(x^2-x+1) - 6 = 0 \), we can let \( f(x) = x(x-1)(x^2-x+1) - 6 \) and find the roots by evaluating \( f(x) \).

### Step 1: Rewrite the Equation
The equation to solve is:
\[
f(x) = x(x-1)(x^2-x+1) - 6 = 0
\]

### Step 2: Expand \( x(x-1)(x^2-x+1) \)
First, we can expand \( x(x-1)(x^2-x+1) \):
- The expression can be computed as follows:
\[
x(x-1) = x^2 - x
\]
Then multiply it by \( (x^2 - x + 1) \):
\[
(x^2 - x)(x^2 - x + 1) = x^2(x^2-x+1) - x(x^2-x+1)
\]
\[
= x^4 - x^3 + x^2 - (x^3 - x^2 + x)
\]
\[
= x^4 - 2x^3 + 2x^2 - x
\]

Now, substituting back into \( f(x) \):
\[
f(x) = x^4 - 2x^3 + 2x^2 - x - 6
\]

### Step 3: Solve \( f(x) = 0 \)
Now we need to solve:
\[
x^4 - 2x^3 + 2x^2 - x - 6 = 0
\]

### Step 4: Factor or Use Numerical Methods
This quartic equation can potentially be solved by factorization or numerical methods, like using the Rational Root Theorem, synthetic division, or numerical approximation methods (like Newton's method or graphing).

### Step 5: Possible Rational Roots
Check possible rational roots (±1, ±2, ±3, ±6). Testing these may reveal any simple roots or give insights for further factorization.

#### Example of Testing a Root:
- Testing \( x = 2 \):
\[
f(2) = 2^4 - 2(2^3) + 2(2^2) - 2 - 6 = 16 - 16 + 8 - 2 - 6 = 0
\]
\( x = 2 \) is a root.

### Step 6: Factor the Polynomial
Now, factor:
\[
f(x) = (x - 2)(x^3 + ax^2 + bx + c)
\]

After factoring out \( x - 2 \), perform polynomial long division or synthetic division to find the cubic polynomial.

### Conclusion
From this cubic polynomial, continue solving for the remaining roots either by factoring, using the cubic formula, or numerical solutions.

This gives a comprehensive approach to solving \( x(x-1)(x^2-x+1) - 6 = 0 \). You can proceed by testing or using graphing calculators to find approximate roots.