Differentiation

x^3 + x^2 - 1>x^3

x^2 - 1>0
x^2 >1

is that wat u mean ?

Edit: @man111:Yeah .. its alright..i had a typo ..which i have corrected..

\hspace{-16}$Let $\mathbf{u=\frac{1}{4}.x^4+\frac{1}{3}.x^3-x}$ and $\mathbf{v=\frac{x^4}{4}}$\\\\ Now Given that $\mathbf{\frac{du}{dx}>\frac{dv}{dx}}$\\\\ $\mathbf{x^3+x^2-1>x^3}$\\\\ $\mathbf{x^2-1>0}$\\\\ $\mathbf{(x-1).(x+1)>0}$\\\\ So $\mathbf{x\in (-\infty\;,-1)\cup (+1\;,\infty)}$

Shubhodip i have not seen your solution.