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1. $(-8)^{4/3}=\bigl((-8){^4\bigr)^{1/3}}=4096^{1/3}=16$.

2. $$ \begin{align*} (-8)^{4/3} &= (-8)^{1+1/3} \\ &= -8\times(-8)^{1/3} \\ &= -8\times (-1)^{1/3}\times 8^{1/3} \\ &= -2\times 8\times (-1)^{1/3} \\ &= -16\times (-1)^{1/3}. \end{align*} $$ So, which is the correct?

Éric Guirbal
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Botond
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  • Welcome to Math.SE. please format expressions. Also what are your thoughts on this problem ? The answer is 16. Why ? – Shailesh Oct 18 '15 at 11:11
  • You will have to specify your definition for $a^b$. It is well-known when $a>0$. But when $a<0$ or $a$ not real at all, you should specify which of the possible definitions you want to use. Unfortunately, there is no definition in these cases with all the nice properties of powers. – GEdgar Oct 18 '15 at 12:43

4 Answers4

1

Both your solutions are correct. To see your first and second solution align note that one solution of $x = (-1)^{1/3}$ is $$x = -1$$

So one possible solution of your original problem is $$-16*(-1)^{1/3} = 16$$

This can be seen by observing that $$(-1)^3 = -1$$

In the complex plane $x= (-8)^{4/3}$ has multiple solutions of the form $-16r_i$ where $r_1,r_2,r_3$ are the complex cube roots of $-1$ which are

\begin{align*} r_1 &= \frac{1}{2} + \frac{\sqrt{3}}{2}i \\ r_2 &= \frac{1}{2} - \frac{\sqrt{3}}{2}i \\ r_3 &= -1 \end{align*}

HBeel
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  • I think $(-1)^{1/3}$ not equals with $-1$, since $(-1)^{1/3}=(e^{iπ})^{1/3}=e^{\frac{iπ}{3}}=cos(\frac{π}{3})+i*sin(\frac{π}{3})=\frac{1}{2}+\frac{3^{1/3}}{2}$ – Botond Oct 18 '15 at 11:45
  • (-1)^(1/3) not equals -1!! – Jan Eerland Oct 18 '15 at 11:49
  • I think you didn't want write $-1^3 = -1$ but $(-1)^3 = -1$. – Éric Guirbal Oct 18 '15 at 11:50
  • @O.Botond You're correct (except you're missing an $i$ in the very last bit), there is 3 solutions of the form $y = x^{1/3}$ (for $y=-1$ the other solutions are the complex conjugate of what you said and $-1$). If you consider $x= (-8)^{4/3}$ in the complex plane there will also be more than one solution obtained by taking the 3 cube roots of $-1$ – HBeel Oct 18 '15 at 11:55
  • @JanEerland see my edit – HBeel Oct 18 '15 at 12:00
  • To define that $(-1)^\frac {1} {3}$ IS -1,we must define that the cuberoot of -1 is -1,which also means$ -1 \times -1 \times -1$ is equals to -1,which in this case,it is true.Don't overcomplicate,I guess – ministic2001 Oct 18 '15 at 12:08
  • @ministic2001 you're right I should be careful with my English, edited to clarify. – HBeel Oct 18 '15 at 12:11
  • Down voters please come forward, I'd be happy to do some further editing – HBeel Oct 18 '15 at 12:16
  • @Hbeel's I don't see much error in your English though because I can understand easily.I'm just a 14 year old kid and a maths enthusiast so that's why I can think that easily,not really learning much about $e^i\pi = -1$ (Not good with Mathjax formatting) lol. – ministic2001 Oct 18 '15 at 12:16
  • @ministic2001 I was more agreeing that I shouldn't define $-1 = (-1)^{1/3}$ which is how it came across due to my choice of words, hence why I changed it to $-1$ being a possible solution to $x = (-1)^{(1/3)}$ – HBeel Oct 18 '15 at 12:18
  • @HBeel I see..... (upvoted) – ministic2001 Oct 18 '15 at 12:20
  • I must downvote, as this is simply inaccurate. $(-8)^\frac{4}{3}$ is not an ambiguous statement, and it doesn't refer to three different numbers. It refers to specifically $-8-8\sqrt{3}i$, as the principal root is always used. E.g. $\sqrt{4}$ is unambiguously $2$, never $-2$. I'll remove my downvote if this is updated. For the same reason, $(-1)^\frac{1}{3}$ is not $-1$, but this has been mentioned above. – Nicholas Pipitone Oct 18 '15 at 15:51
1

Using the following rules, with $a,b \in \mathbb{R}$:

  1. $\left|a^b\right|=\left|a\right|^b$;
  2. $\arg\left(a^b\right)=\tan^{-1}\left(\cos(b\cdot\arg(a)),\sin(b\cdot\arg(a))\right)$

$$(-8)^{\frac{4}{3}}=$$ $$\left|(-8)^{\frac{4}{3}}\right|e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$ $$\left|-8\right|^{\frac{4}{3}}e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$ $$8^{\frac{4}{3}}e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$ $$16e^{\arg\left((-8)^{\frac{4}{3}}\right)i}=$$ $$16e^{\tan^{-1}\left(\cos\left(\frac{4\pi}{3}\right),\sin\left(\frac{4\pi}{3}\right)\right)i}=$$ $$16e^{\tan^{-1}\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right)i}=$$ $$16e^{-\frac{2\pi}{3}i}=$$ $$16\cos\left(-\frac{2\pi}{3}\right)+16\sin\left(-\frac{2\pi}{3}\right)i=$$ $$16\cdot \left(-\frac{1}{2}\right)+\left(16\cdot -\frac{\sqrt{3}}{2}\right)i=$$ $$-8+(-8\sqrt{3})i=$$ $$-8-8\sqrt{3}i$$

So:

$$(-8)^{\frac{4}{3}}=-8-8\sqrt{3}i$$

Jan Eerland
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1

Let $a$ be any real number. We can show that the equation $$ x^3 = a $$ has one and only one real root. It is called the cube root of $a$ and is denoted $a^{1/3}$ or $\sqrt[3]{a}$. From $(-1)^3 = -1$, we conclude that $(-1)^{1/3} = -1$.

Éric Guirbal
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1

They are both correct.

$$(-1)^{3} = -1$$

because it is

$$-1 \times -1 \times -1 $$

and a negative $\times$ a negative is a positive:

\begin{align*} (-1 \times -1) \times -1 \\ = 1 \times -1 \\ = -1 \\ \end{align*}

Because of that, a solution to "What is the cube root of -1 ($\sqrt[3]{-1}$)" is $-1$.

This means that $$-16 \times -1^{1/3} = 16$$

can also be written as

$$-16 \times -1 = 16$$

which is clearly true.

Also, it may be easier to solve

$$−8^{4/3}$$

with the following method:

\begin{align*} -8^{4/3} \\ &=\sqrt[3]{-8}^4 \\ &=2^4 \\ &= 2 \times 2 \times 2 \times 2 \\ &=16 \\ \end{align*}

Tim
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