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 ?
 A: Consider that you look for the zero of function
$$f(x)=x^3-4$$
We have $f(1)=-3$ and $f(2)=+4$. Using $x_0=\frac 32$ (the midpoint), the first iterate of Newton method is $\frac{43}{27} \approx 1.59259$ while the exact value is  $1.58740$.
The answer is then $1.59$.
Edit
You do not need to know about Newton method since what I did is just linearization (or Taylor series)
$$f(x)=f(x_0)+f'(x_0) (x-x_0)+ O((x-x_0)^2$$
so for you case
$$0=-\frac 58+\frac {27}4 (x-\frac 32)=\frac{27 }{4}x-\frac{43}{4}\implies x=\frac{43}{27}$$
A: If you are around powers of 2 a lot (as computer scientists are, for example), then you can recognize $4.096=2^{12}/10^3=(2^4/10)^3=1.6^3$, and then $4^{1/3} = 1.6(1-\frac{0.096}{4.096})^{1/3}$ is much better approximated by its power series. Indeed, the approximation $1.6(1-\frac13\frac{0.096}{4.096}) \approx 1.6(1-\frac13\frac{0.1}4) = 1.6 - \frac1{75}$ is already within $0.05\%$ of the true answer $\approx 1.5874$.
A: How about using some Calculus? We got to compute $\sqrt[3]{4}$, so let $y=f(x) =\sqrt[3]{x}$. Now, we know that $f(8)= 2$ and
$$
\frac{dy}{dx}= \frac{1}{3} \frac{1}{x^{2/3}}$$
$$\Delta x = 4 - 8 = -4 \\
\Delta y = 1/3 \frac{1}{x^{2/3}} \Delta x \\
\Delta y = \frac{1}{3 \times 4} (-4) \\
\Delta y = -0.3333 
$$
So, $f(4) = 2 - 0.3333= 1.667$.
I know my answer is very far from original one, and to be honest it looks to me that they have used the calculator for the correct answer, I added this method because you've mentioned the time limit and wanted something quick.
A: 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.
A: $\displaystyle\,\sqrt[\large 3]{\, 4\, } =
\sqrt[\large 3]{512 - 12 \over 125} =
{8 \over 5}\left(1 - {3 \over 128}\right)^{1/3} \approx
1.6 \times \left(1 - {1 \over 128}\right) = 1.587\color{red}{5}$

In general ( expanding the above binomial ), with
$\displaystyle{8 \over 5}\sum_{k = 0}^{n}{1/3 \choose k}
\left(\,{-\,{1 \over 128}}\,\right)^{k}$:
$$
{\Large\,\sqrt[\large 3]{\, 4\, }\,}\ \approx\ \left\{\begin{array}{lll}
{\displaystyle 1.\color{red}{6}} & {\displaystyle n = } & {\displaystyle 0}
\\
{\displaystyle 1.587\color{red}{5\ldots}} & {\displaystyle n = } & {\displaystyle 1} 
\\
{\displaystyle 1.58740\color{red}{2\ldots}} & {\displaystyle n = } & {\displaystyle 2}
\\
{\displaystyle 1.5874010\color{red}{7\ldots}} & {\displaystyle n = } & {\displaystyle 3}
\\
{\displaystyle 1.58740105\color{red}{2\ldots}} & {\displaystyle n = } & {\displaystyle 4}
\\
{\displaystyle 1.5874010519\color{red}{7\ldots}} & {\displaystyle n = } & {\displaystyle 5}
\\
{\displaystyle 1.587401051968\color{red}{3\ldots}} & {\displaystyle n = } & {\displaystyle 6} 
\end{array}\right.
$$

Result for some machine software:
$$
1.587401051968199474751705639272308260391493327899853009808\ldots
$$
