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

(A) Given function is $f(x) = x^{3} - 3x^{2} + 3x - 100$.
To check if the function is increasing,we find its derivative $f'(x)$ with respect to $x$:
$f'(x) = \frac{d}{dx}(x^{3} - 3x^{2} + 3x - 100)$
$f'(x) = 3x^{2} - 6x + 3$
Factor out $3$ from the expression:
$f'(x) = 3(x^{2} - 2x + 1)$
Recognize the perfect square trinomial:
$f'(x) = 3(x - 1)^{2}$
Since $(x - 1)^{2} \geq 0$ for all $x \in R$,it follows that $f'(x) = 3(x - 1)^{2} \geq 0$ for all $x \in R$.
Since the derivative $f'(x)$ is non-negative for all $x \in R$,the function $f(x)$ is an increasing function in $R$.