Why does cube rooting a negative number always give us an answer?

Question

Well, I multiply a negative number twice by itself and I always get a negative number! Look:$$-1.5\cdot-1.5\cdot-1.5=-3.375$$$$-6\cdot-6\cdot-6=-216$$I think this is why. I also know that the number rooting the number inside the radical sign is odd. This must also be why. At least I know that the cube root of $-8$ is $-2$ because $-8/-2=4/-2=-2$. This must be a third thing why this can happen. Do you agree?

Answer 1

The cube root $y$ of any number $x$ is the unique number such that $y^3 = x$. If $x$ is positive, it makes sense that we can find a positive number $y$ such that $y \cdot y \cdot y =y^3 = x$ because a product of three positive numbers is positive. If $x$ is negative, it also makes sense that we can find some negative $y$ such that $y^3=x$, because the product of three negative numbers $(y \cdot y \cdot y)$ will also be negative. As you most likely know, if $x$ is negative, it will not have a real square root. This is because for all real numbers $y$ (positive or negative), $y^2$ is always positive.

Answer 2

Let us make the assumption that multiplying two positive numbers gives you a positive number.

Suppose by contradiction that there exists $n< 0$ such that $m = \sqrt[3]{n}> 0$. Then we will have $$m^2=\underbrace{m}_{>0}\cdot\underbrace{m}_{>0} >0,$$ and thus $$m^3 = \underbrace{m^2}_{>0} \cdot \underbrace{m}_{>0}>0,$$ which is a contradiction with $m^3= n<0$.