We know $$-1<\mathrm{x}<0$$; 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).