QPaperGen

Let
\[
f(x) = x^3 - 3x^2 + 3x
\]

Differentiate \( f(x) \) with respect to \( x \):
\[
f'(x) = \frac{d}{dx}(x^3 - 3x^2 + 3x) = 3x^2 - 6x + 3
\]

Factor the derivative:
\[
f'(x) = 3x^2 - 6x + 3 = 3(x^2 - 2x + 1) = 3(x - 1)^2
\]

Now, for all \( x \in \mathbb{R} \), we know:
\[
(x - 1)^2 \geq 0 \Rightarrow f'(x) = 3(x - 1)^2 \geq 0
\]

Also, \( f'(x) = 0 \) only at \( x = 1 \), and \( f'(x) > 0 \) elsewhere.
Thus, \( f'(x) \geq 0 \) for all \( x \in \mathbb{R} \), and \( f'(x) > 0 \) for all \( x \ne 1 \).
\[\]
\(\textbf{Conclusion:}\)
\(
f(x) = x^3 - 3x^2 + 3x \text{ is an increasing function on } \mathbb{R}.
\)