Prove that the function given by $f\left(x\right) = x^3 - 3x^2 +3x - 100$ is increasing in $\mathbb{R}$.
$f\left(x\right) = x^3 -3x^2 + 3x -100$
$\begin{aligned}
\therefore f'\left(x\right) & = 3x^2 -6x +3 \\
& = 3 \left(x^2 -2x +1\right) \\
& = 3 \left(x-1\right)^2
\end{aligned}$
Now, $\left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$
$\therefore$ $3 \left(x-1\right)^2 > 0 \; \forall \; \mathbb{R}$
$\implies$ $f'\left(x\right) > 0 \; \forall \; \mathbb{R}$
$\implies$ $f\left(x\right)$ is strictly increasing in $\mathbb{R}$.