$\frac{\mathrm{d} (x^3+bx^2+cx+d)}{\mathrm{d} x}= (3x^2+2bx+c)$
$x=\frac{\mathrm{b^{2}\pm \sqrt{b^2-4c}} }{\mathrm{2} }$
$b^{2}-4c$<0 (0 <$b^{2}$< c)
$(3x^2+2bx+c)>0 $ Hence Increasing
there's way to know this...