1. If the constant term in the expansion of $\left(\frac{1}{x}-x^{2}\right)^{n}$ is 15, then the coefficient of $x^{3}$ is
1. -20 .

Let the $(r+1)$-th term be the constant term, then $\mathrm{C}_{n}^{r}\left(\frac{1}{x}\right)^{n-r} \cdot\left(-x^{2}\right)^{r}=(-1)^{r} \mathrm{C}_{n}^{r} x^{-n+3 r}$, from $-n+3 r=0$ we get $n=3 r$. Also, $\mathrm{C}_{n}^{r}=15$, so $n=6, r=2$. Let the $(m+1)$-th term $(-1)^{m} \mathrm{C}_{6}^{m} x^{-6+3 m}$ be the term containing $x^{3}$, then $-6+3 m=3 \Rightarrow m=3$. Therefore, the coefficient of $x^{3}$ in the expansion is $(-1)^{3} \mathrm{C}_{6}^{3}=-20$.