1. Function
$$
f(x)=27^{x}-3^{x+3}+1
$$

The minimum value of the function on the interval $[0,3]$ is $\qquad$
- 1. -53.

Let $t=3^{x}(x \in[0,3])$. Then $f(x)=g(t)=t^{3}-27 t+1(t \in[0,27])$. And $g^{\prime}(t)=3 t^{2}-27=3(t-3)(t+3)$, so when $t \in[1,3]$, $g^{\prime}(t)0, g(t)$ is monotonically increasing.
Therefore, when $t=3$, $g(t)$ reaches its minimum value
$$
g(t)_{\min }=g(3)=-53,
$$

which means when $x=1$, $f(x)$ takes the minimum value -53.