Difference between $(-x)^{1/n}$ and $-x^{1/n}$?

Question

I know these seems to be splitting hairs, but according to two different calculators, there's a real difference.

For e.g, $(-8)^{1/3}$ computes to NaN, where $-8^{1/3}$ computes to $-2$.

Though am having a hard time seeing what the brackets actually mean, I mean, in this context at least. They seem arbitrary, though is there any actual use for the former notation?

By the way, this came about when trying to compute the more complex expression, $(3/-96)^{1/5}$, where of course, you have to bracket the $3/-96$, to tell the calc you want it to raise both numbers to said power.

Answer 1

Well, there's an order of operations you have to take into account. Performing $-8^{1/3}$ should translate to doing $$-\left(8^{1/3}\right)=-(2)=-2,$$

while $(-8)^{1/3}$ translates to taking a cuberoot of a negative number first. Now, taking the cuberoot of a negative number is certainly possible, but for some reason your calculator doesn't like it.

Note that there are technically multiple solutions to both $x^3=8$ and $x^3=-8$, but there is only one real solution in each case, and generally the real solution of $x=2$ and $x=-2$ respectively are thought of as the cuberoots.

Now, as far as your actual question, note that $$\left(-\frac{3}{96}\right)^{1/5}=\left(-\frac{1}{32}\right)^{1/5}$$ This can be evaluated without a calculator.

Answer 2

In English, "$(-x)^{1/n}$" is the "$n$-th root of negative $x$", whereas "$-x^{1/n}$" is the "negative $n$-th root of $x$".

The difference is in the order of operations. In "$(-x)^{1/n}$", one negates $x$, then takes the $n$-th root, whereas in "$-x^{1/n}$", one takes the $n$-th root of $x$, then negates the result.

Clear?

Answer 3

Normally rational exponents are reserved toh positive numbers because with negative numbers, this notation can lead to inconsistencies: $$-2=(-8)^{\tfrac13}=(-8)^{\tfrac26}=\begin{cases}\bigl((-8)^2\bigr)^{\tfrac16}=(64)^{\tfrac16}=2,\\\Bigl((-8)^{\tfrac16}\Bigr)^2\enspace\text{which is not defined}.\end{cases}$$ Of courseone may speak of the cube root of a negative number, but this is to be denoted $\sqrt[3]{x}$, not $x^{\tfrac13}$.

One more argument against the use of rational exponents for negative numbers: the definition is: $$x^{\tfrac1n}\stackrel{\text{def}}{=}\mathrm e^{n\ln x},$$ which definitely excludes non-positive numbers.