Prove that the function given by $f(x) = {x^3} - 3{x^2} + 3x - 100$ is increasing in $R$.

We have, $f(x) = {x^3} - 3{x^2} + 3x - 100$ …(i)

Differentiating (i) w.r.t. $x$, we get
$f'(x) = 3{x^2} - 6x + 3 = 3({x^2} - 2x + 1) = 3{(x - 1)^2} \ge 0$ for all $x \in R$

[$3 > 0,{(x - 1)^2} \ge 0$ (being perfect square)]

$\Rightarrow f(x) \ge 0 \Rightarrow f(x)$ is increasing on $R$