Answer
All real numbers
Work Step by Step
The domain of $f(t)=\sqrt[3] {t-1}$ is the set of all real numbers $t$ for which $\sqrt[3]{t-1}$ is defined as a real number. We know we can take cube roots of negative numbers, zero, and positive numbers, so the domain of $f(t)=\sqrt[3] {t-1}$ is all real numbers $t$ for which $t-1$ is a real number. But $t-1$ is a real number for all real numbers $t$. Hence the domain of our function is all real numbers.
In set notation, the domain is $\{t:t\in R\}= \mathbb{R}.$
In interval notation, the domain is $(-\infty , \infty).$
