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I am comfuse about something.

I want to compute $(-8)^\frac{2}{3}$

Is it $(-2^3)^\frac{2}{3}$=$(-2)^{3\cdot\frac{2}{3}}$=$(-2)^2=4$ ?

Is there any problem here because the base is negative?

Thanks.

Hippalectryon
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ruth
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5 Answers5

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$$(-8)^{\dfrac23}=\left[(-8)^2\right]^{\dfrac13}=(64)^{\dfrac13}=(4^3)^{\dfrac13}$$

$=4\cdot w$ where $w$ is a cube root of unity

lab bhattacharjee
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  • isn't the answer is just $4$? or there are more answers (complex) – ruth Oct 03 '14 at 06:39
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    @ruth, If $x=\sqrt[3]{64},x^3=64$ and http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html says "a polynomial of degree $n$ has $n$ root" – lab bhattacharjee Oct 03 '14 at 06:42
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    @ruth, See also: http://en.wikipedia.org/wiki/Principal_value#Complex_argument and http://en.wikipedia.org/wiki/Principal_branch – lab bhattacharjee Oct 03 '14 at 06:45
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    ok but here it's not $X^3=64$ but $x=\sqrt[3]{64}$ so there is just one value which is $4$ i cant see why it's $4\cdot w$. in your link they say that $w=\frac{-1+\sqrt3i}{2}$ – ruth Oct 03 '14 at 06:47
  • @ruth, $w$ can be any one of all the three values – lab bhattacharjee Oct 03 '14 at 06:48
  • so all other answers here are incomplite. – ruth Oct 03 '14 at 06:49
  • @ruth, All the three values are legitimate – lab bhattacharjee Oct 03 '14 at 06:50
  • i would glad if you could see why i got the right answer if my transitions were wrong. – ruth Oct 03 '14 at 07:42
  • @labbhattacharjee. I think there is a point you should discuss here because I think it is very imoprtant. If you input the expression in WA or other CAS, you get $\cdots$ $-2+2 i \sqrt{3}$ – Claude Leibovici Oct 03 '14 at 07:47
  • @ClaudeLeibovici, I think the principal value of $\sqrt[3]1=1$. I'm not sure why CAS or WA say that. – lab bhattacharjee Oct 03 '14 at 08:32
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    @labbhattacharjee. This is a very common situation. In fact, I heard that Mathematica contains now a cube root function. – Claude Leibovici Oct 03 '14 at 08:46
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    I don't agree that $64^{1/3}$ has three values. The cube root of a real number is understood to be a real number, unless there is a specific reason to think otherwise. (If not, where do you stop? What about quaternions, octonions,...?) – TonyK Oct 04 '14 at 21:53
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    For being the accepted answer, I find this very badly stated and confusing. You seem to be saying that the original expression does not have one value, which means it is not well defined; this should be made clear. Also applying the law of exponents with negative base is well known to be not allowed, at the risk of getting contradictions. Finally $64^{1/3}$ means $\sqrt[3]{64}$ (certainly at the maturity level of OP) which does have a well defined value. I vote $(-1)^{6/2}$. – Marc van Leeuwen Oct 05 '14 at 05:15
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Your answer is correct.

$(-8)^{2/3}=\left((-8)^{1/3}\right)^2=(-2)^2=4$

$(-8)^{2/3}=\left((-8)^2\right)^{1/3}=(64)^{1/3}=4$

JRN
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Please have a look at: How do you compute negative numbers to fractional powers?

With rational exponents for negative numbers you are in trouble. You can do $x^{1/3}$ but if you try the thing you want then there are problems. You find that it depends on the order how you perform the operations resp. which root you choose. Therefore if we consider only real numbers, such things as $(-8)^{2/3}$ are not well defined. See also: http://en.wikipedia.org/wiki/Exponentiation#Rational_exponents

here is the example from this site: $−27 = (−27)^{((2/3)⋅(3/2))} = ((−27)^{2/3})^{3/2} = 9^{3/2} = 27$

Community
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Karl
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  • In general, powering then root extraction is not commutative with root extraction then powering. It matters which operation is done first.

    However, positive real bases are a special case for which this is commutative. When the root being extracted is odd, it is commutative for all real numbers.

    I explain here.

    https://math.stackexchange.com/a/4864176/928654

     – Michael Ejercito Feb 16 '24 at 19:07
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Yes. You are right. Or you can trace in this way:

$=\sqrt[3]{(-8)^2}=\sqrt[3]{64}=4$

Paul
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As $2$ and $3$ are coprime integers, riaising to the ($\frac23$)th power (equivalently, taking the ($\frac32$)th root, is esdsentially squaring and cubing.

While generally, using the simple $p$th power then $q$th root to calculate the ($\frac pq$)th power generally works for only positive real bases, it works in the case of negative numbers for a $(\frac23$)th power.

$$\sqrt[3]{(-8)^2}=\sqrt[3]64=2$$

$$(\sqrt[3]{(-8)})^2=(-2)^2=2 $$

This does not seem to work for all exponents.

$$\sqrt{(-1)^3}=\sqrt{-1}=i$$

if $i^2=-1$

However,

$$(\sqrt{-1})^3=i^3=-i\neq-1$$

But why?

Let us look at the numerator and denominator.

We were taking a square root, which is an even root, in this example. In the $(-8)^{\frac23}$ example, we were taking a cube root, which is an odd root.

Anbd here we have it. For all nonzero integers $z$, none of the $2z$th roots of negative real numbers are real numbers.

But for a ($2z+1)$th root, every real number, including negative numbers, has a unique real root. It follows that for integers $z_1$ and $z_2$, and negative real numbers $-r$,

$$(-r)^{\frac{z_1}{2z_2+1}}=\sqrt[\frac{2z_2+1}{z_1}]{-r}=\sqrt[2z_2+1]{(-r)^{z_1}}=(\sqrt[2z_2+1]{(-r)})^{z_1}$$

has a unique real value, and powering then root extraction yields the same value as root extraction then powering.

Michael Ejercito
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