Let $$ f(x)=x^{2025}-x^{2000}, x \text{ }\epsilon \text{ }[0,1] $$ and the minimmu value of the function $$ f(x)$$ in the interval [0, 1] be $$(80)^{80}(n)^{-81}$$. Then n is equal to

To find the minimum of $$f(x)=x^{2025}-x^{2000}$$ on $$[0,1]$$, we first compute its derivative.

Derivative formula: If $$g(x)=x^n$$, then $$g'(x)=n\,x^{n-1}$$.

Applying to $$f$$, we get
$$f'(x)=2025\,x^{2024}-2000\,x^{1999}$$. To find critical points, set $$f'(x)=0$$:

$$2025\,x^{2024}-2000\,x^{1999}=0$$

Factor out common term $$x^{1999}$$:

$$x^{1999}\bigl(2025\,x^{25}-2000\bigr)=0$$. Thus $$x^{1999}=0$$ or $$2025\,x^{25}-2000=0$$.

From $$x^{1999}=0$$, we get $$x=0$$. From $$2025\,x^{25}=2000$$, we get

$$x^{25}=\frac{2000}{2025}=\frac{80\cdot25}{81\cdot25}=\frac{80}{81}$$.

Thus the critical point in $$(0,1)$$ is
$$x=\Bigl(\tfrac{80}{81}\Bigr)^{1/25}\,.$$

Evaluate $$f$$ at the endpoints and this critical point:

At $$x=0$$, $$f(0)=0^{2025}-0^{2000}=0\,$$.

At $$x=1$$, $$f(1)=1^{2025}-1^{2000}=1-1=0\,$$.

At $$x=\Bigl(\tfrac{80}{81}\Bigr)^{1/25}$$, note that $$x^{25}=\tfrac{80}{81}$$, so

$$x^{2000}=\bigl(x^{25}\bigr)^{80}=\Bigl(\tfrac{80}{81}\Bigr)^{80},\quad x^{2025}=\bigl(x^{25}\bigr)^{81}=\Bigl(\tfrac{80}{81}\Bigr)^{81}.$$

Hence

$$f(x)=x^{2025}-x^{2000} =\Bigl(\tfrac{80}{81}\Bigr)^{81} -\Bigl(\tfrac{80}{81}\Bigr)^{80} =\Bigl(\tfrac{80}{81}\Bigr)^{80}\Bigl(\tfrac{80}{81}-1\Bigr) =-\frac{80^{80}}{81^{81}}\,.$$

The minimum value among $$0,\,0,\,-\frac{80^{80}}{81^{81}}$$ is $$-\frac{80^{80}}{81^{81}}$$.

We are given that the minimum value equals $$(80)^{80}(n)^{-81}$$. Therefore,

$$ (80)^{80}(n)^{-81} =-\frac{80^{80}}{81^{81}}\,. $$

Canceling $$(80)^{80}$$ from both sides yields

$$ n^{-81}=-\frac{1}{81^{81}}\,. $$

For $$n=-81$$,

$$n^{-81}=(-81)^{-81}=\frac{1}{(-81)^{81}} =-\frac{1}{81^{81}}\,, $$

so

$$ (80)^{80}(n)^{-81} =(80)^{80}\Bigl(-\tfrac{1}{81^{81}}\Bigr) =-\frac{80^{80}}{81^{81}}\,, $$

which matches the minimum value. Therefore, $$n=-81$$.

Answer: Option D