evaluate the expression.\n$sqrt3{-1}$

Explanation:

Step1: Recall cube - root definition

The cube - root of a number $x$, denoted as $\sqrt[3]{x}$, is a number $y$ such that $y^3=x$. We want to find $y$ where $y^3=-1$.

Step2: Consider the equation $y^3 + 1=0$

We know that $a^3 + b^3=(a + b)(a^2 - ab + b^2)$. So, $y^3+1=(y + 1)(y^2 - y+1)=0$. Setting $y + 1=0$, we get $y=-1$. Setting $y^2 - y + 1=0$, using the quadratic formula $y=\frac{1\pm\sqrt{1 - 4}}{2}=\frac{1\pm\sqrt{- 3}}{2}=\frac{1\pm i\sqrt{3}}{2}$ for the complex - valued roots. But for the principal cube - root of $-1$, we consider the real root.

Answer:

$-1$