Simplifying $y=2^{2/3} + 2^{-1/3}$ I am working on a calculus problem where I have to find the local minimum. The value I got was $$y=2^{2/3} + 2^{-1/3}.$$ I simplified it and got this:
$$ y=2^{2/3} + \frac{1}{2^{1/3}}$$
$$y=\frac{2^{2/3}2^{1/3}}{2^{1/3}} + \frac{1}{2^{1/3}}\frac{2+1}{2^{1/3}}=\frac{3}{2^{1/3}}$$
According to the online homework and Wolfram Alpha, the correct answer is :
$$\frac{3}{2^{2/3}}$$
Did I miss a step while simplifying the answer? I don't understand why it is $2/3$ instead of $1/3$.

Your simplification is entirely correct.  Using your original evaluation of $y$, $$y=e^{2\frac{\ln2}{3}}+e^{-\frac{\ln2}{3}}=2^{2/3} + 2^{-1/3} = \dfrac{3}{2^{1/3}}$$
Are you sure your value for $y$ is a correct mimimum?
EDIT:
Original function: $$y(x)=e^{2x}+e^{-x}$$
Critical point (minimum) at $x = \dfrac{-\ln 2}{3} \implies$
$$y=e^{-2\frac{\ln2}{3}}+e^{\frac{\ln2}{3}}=2^{-2/3} + 2^{1/3} = \dfrac{3}{2^{2/3}}$$
