We know $\(-1<x<0\)$; we need to check that which of the given options is greatest.

As x is negative the value of $\(\mathrm{x}^{\text {odd \)$ will also be negative and also as x lies between -1 and 0 , the higher odd power will result in higher negative value.

Out of the given options (A) \& (B) will be negative so cannot be the greatest, so the comparison remains among the remaining three options.

We know $\(\mathrm{x}^3>\mathrm{x}\)$ but $\(-\mathrm{x}^3<-\mathrm{x} \Rightarrow 1-\mathrm{x}^3<1-\mathrm{x}\)$, so option (C) is greater than option (D).

Finally we need to compare option (C) with option (E).

$\(\mathrm{x}^4>\mathrm{x}\)$ but $\(-\mathrm{x}^4<-\mathrm{x} \Rightarrow 1-\mathrm{x}^4<1-\mathrm{x}\)$ which means $\(1-\mathrm{x}\)$ is the greatest.


Hence the answer is (C).