Better methods to approximate $2^{2\over 3}$ Recently while solving a problem on thermodynamics I ended up with $2^{2\over 3}$ .
Now the problem was on a test where no calculators were allowed and answer was required upto $2$ decimal digits.
I then resorted to binomial theorem for help (for $x\lt 1$) $$\left. \begin{array} { l } { ( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) x ^ { 2 } } { 2 ! } + \frac { n ( n - 1 ) ( n - 2 ) } { 3 ! } x ^ { 3 } + \ldots \ldots + \frac { n ( n - 1 ) \ldots \ldots ( n - r + 1 ) } { r ! } x ^ { r } \ldots  } \\  \end{array} \right.\text{upto}\, \,  \infty$$
So the original problem can be written as :
$$2^{2\over 3}=4^{1\over 3}=(8-4)^{1\over 3}=2\left(1-\frac{1}{2}\right)^{1\over 3}$$
Now after evaluation first $3$ terms I ended up with $\left(2-\frac{1}{3}-\frac{1}{18}\right) \approx 1.61$ but the correct answer was $1.59$.
Also the average time you got per question was around $2$ minutes and I had already used more than half of it so I did not think of calculating more terms.
Now I am looking for a method which can help me evaluate $2^{2\over 3}$ faster and more precisely.
Can anyone please help me with this ?

Perhaps more elaboration on how to use Newton's method, since I personally find it very easy to use to find the first few digits of $n$th roots using only basic calculations.
From the binomial expansion, we know that
$$(x+\Delta x)^3=x^3+3x^2\Delta x+\mathcal O((\Delta x)^2)$$
Our goal is essentially finding the $\Delta x$ that fixes $x^3$ so that $(x+\Delta x)^3$ is closer to $4$. In this way, we can find $\sqrt[3]4$.
Setting it equal and solving, we end up with
$$x^3+3x^2\Delta x=4\implies\Delta x=\frac{4-x^3}{3x^2}=\dfrac{\dfrac4{x^2}-x}3$$
Adding $x$ to both sides to find our new $x$ gives us
$$x+\Delta x=\frac{4+2x^3}{3x^2}=\dfrac{\dfrac4{x^2}+2x}3$$
Starting with values of $x$ closer to $\sqrt[3]4$ such as the $1.61$ you found will lead to faster correction, but for example purposes we start far away at $x=1$, which makes the next estimate become
$$x+\Delta x=\frac{4+2}3=2$$
Substituting $x=2$ in now gives us
$$x+\Delta x=\frac{4+16}{12}=\frac53\simeq1.66$$
The next estimate is a tad messier but still manageable:
$$x+\Delta x=\frac{4+2\times\frac{125}{27}}{3\times\frac{25}9}=\frac{4\times27+2\times125}{3\times3\times25}=\frac{108+250}{225}=\frac{358}{225}\simeq1.591$$

In general you will find that
$$x+\Delta x=\frac{a+(n-1)x^n}{nx^{n-1}}$$
for computing $\sqrt[n]a$, which works out to give the first few digits fairly fast and easily provided a good initial estimate (as you had done) and that $n$ is not too large.
