Easiest way to solve the equation $x^\frac 43 = \frac {16}{81}$ What would be the easiest way to solve this?
$$x^\frac 43 = \frac {16}{81}$$
I saw this in class and have no clue how did they get $$x = \frac 8{27}$$ 

Method 1:  We transform the equation into a difference of fourth powers, then factor to find the roots of the quartic.
\begin{align*}
x^{\frac{4}{3}} & = \frac{16}{81}\\
x^4 & = \left(\frac{16}{81}\right)^3\\
x^4 & = \left(\frac{2^4}{3^4}\right)^3\\
x^4 & = \frac{2^{12}}{3^{12}}\\
x^4 & = \left(\frac{2^3}{3^3}\right)^4\\
x^4 & = \left(\frac{8}{27}\right)^4\\
x^4 - \left(\frac{8}{27}\right)^4 & = 0\\
\left[x^2 + \left(\frac{8}{27}\right)^2\right]\left[x^2 - \left(\frac{8}{27}\right)^2\right] & = 0\\
\left[x^2 + \left(\frac{8}{27}\right)^2\right]\left(x + \frac{8}{27}\right)\left(x - \frac{8}{27}\right) & = 0 \tag{1}
\end{align*}
For any real number $x$, the factor
$$x^2 + \left(\frac{8}{27}\right)^2 > 0$$
so the real-valued roots of equation 1 are 
$$x = \pm \frac{8}{27}$$
Note:  If $x$ is a non-zero real number, then
$$
(x^n)^{\frac{1}{n}} = 
\begin{cases}
x & \text{if $n$ is odd}\\
|x| & \text{if $n$ is even}
\end{cases}
$$
Thus, if $x$ is a non-zero real number, we obtain 
$$\left(x^{\frac{4}{3}}\right)^{\frac{3}{4}} = \left\{\left[\left(x^{\frac{1}{3}}\right)^4\right]^3\right\}^{\frac{1}{4}} = \left\{\left[\left(x^{\frac{1}{3}}\right)^3\right]^4\right\}^{\frac{1}{4}} = (x^4)^{\frac{1}{4}} = |x|$$
Method 2: We raise each side of the equation $x^{\frac{4}{3}} = \frac{16}{81}$ to the $(3/4)$th power, then solve for the real-valued roots.
\begin{align*}
x^{\frac{4}{3}} & = \frac{16}{81}\\
\left(x^{\frac{4}{3}}\right)^{\frac{3}{4}} & = \left(\frac{16}{81}\right)^{\frac{3}{4}}\\
|x| & = \left[\left(\frac{16}{81}\right)^{\frac{1}{4}}\right]^3\\
|x| & = \left[\left(\frac{2^4}{3^4}\right)^{\frac{1}{4}}\right]^3\\
|x| & = \left(\frac{2}{3}\right)^3\\
|x| & = \frac{8}{27}\\
x & = \pm \frac{8}{27}
\end{align*}
Check: If $x = \pm \frac{8}{27}$, then
$$\left(\pm \frac{8}{27}\right)^{\frac{4}{3}} = \left[\left(\pm \frac{8}{27}\right)^{\frac{1}{3}}\right]^4 = \left[\left(\pm \frac{2^3}{3^3}\right)^\frac{1}{3}\right]^4 = \left(\pm \frac{2}{3}\right)^4 = \frac{16}{81}$$
