(-8)^(4/3) is equals with 16 or (-16)*(-1)^(1/3)? 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?
 A: 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*}
A: Using the following rules, with $a,b \in \mathbb{R}$:


*

*$\left|a^b\right|=\left|a\right|^b$;

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


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