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$\frac{\mathrm{d} (x^3+bx^2+cx+d)}{\mathrm{d} x}= (3x^2+2bx+c)$

$x=\frac{\mathrm{b^{2}\pm \sqrt{b^2-4c}} }{\mathrm{2} }$   

$b^{2}-4c$<0     (0 <$b^{2}$< c)

$(3x^2+2bx+c)>0   $ Hence  Increasing

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