If $f(x)=x^3+b x^2+c x+d$ and $0<b^2<c$, then in $(-\infty, \infty)$

f(x) is strictly increasing function

We have,

$f(x)=x^3+b x^2+c x+d \Rightarrow f^{\prime}(x)=3 x^2+2 b x+c$

Let $D_1$ be the discriminant of $f^{\prime}(x)=3 x^2+2 b x+c$. Then,

$D_1=4 b^2-12 c=4\left(b^2-c\right)-8 c<0 \quad\left[∵ b^2<c \text { and } c>0\right]$

$f^{\prime}(x)>0$ for all $x \in(-\infty, \infty)$

f(x) is strictly increasing function on $(-\infty, \infty)$