<s>[INST] (-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?
 [/INST] 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*}
</s>