How to find a cube root of numbers? While, I was solving a problem of Chemistry [Solid State] when I encountered an equation like : $$a^3 = 3.612 \times 10^{-23} $$ 
Where, a is just a quantity [Actually, it is the length of a cubic unit cell.]
I was required to find the cube-root of $a^3$ as I had to use $a$ to find another quantity which was actually my answer.
I know that the cube-root of $a^3$ here can be found by using logarithm. 
$$\begin{align} & \log{a^3} = \log{(3.612 \times 10^{-23})} \\
& 3\log{a} = \log{(36.12 \times 10^{-24})} \\
& 3\log{a} = \log{(36.12)} + \log{(10^{-24})} \\
& 3\log{a} = \log{(36.12)} - 24 \\
& \log{a} = \cfrac{\log{(36.12)} - 24}{3} \end{align} $$
By using logarithm table, I found $\log{(36.12)} = 1.557$ 
Now, just putting this : $$\begin{align} & \log{a}  = \cfrac{1.557 - 24}{3}  \\ & \\ & \log{a} = \cfrac{-22.443}{3} \end{align} $$
Now, taking antilog both sides, this becomes : $$a = antilog(-7.481) $$ 
Now, by using anti-log table, I got this as $$\boxed{a = 3.303 \times 10^{-8} } $$
Now, as you all might have noticed that this is a lot longer method and this has larger probability of a student doing a mistake in the calculations etc.
Is there a shorter or any proper method to find cube-root of a number like stated above?  
 A: Given a log/antilog table and no calculator, your approach is probably optimal. Without such a table, I think you are better off proceeding as:
$$\left ( 3.612 \cdot 10^{-23} \right )^{1/3} \\
= \left ( 36.12 \cdot 10^{-24} \right )^{1/3} \\
= 36.12^{1/3} \cdot 10^{-8}$$
Then do bisection to find $36.12^{1/3}$. Note that $3^3=27<36.12$ and $4^3=64>36.12$. (If you can't find a small interval quickly, just use $[1,10]$.) So now you keep checking midpoints until you get your desired level of precision: $3.5^3 > 36.12$, $3.25^3<36.12$, $3.375^3>36.12$, etc. If at some point the arithmetic gets to be too messy, you can round, at the expense of some convergence speed. For example, you could check $3.4$ instead of $3.375$, then $3.4^3>36.12$, so you could check $3.3$. Then $3.3^3<36.12$, so you have $a=(3.35 \pm 0.05) \cdot 10^{-8}$.
You can also do Newton's method, which in this case is $x_{k+1} = \frac{2 x_k}{3} + \frac{c}{3 x_k^2}$ where you want $c^{1/3}$, but this is hard to do by hand, because the fractions quickly become very awkward, even with rounding.
A: I think your method is probably the best way to solve for the cube root of a number.
I found some sites that offer quick solution towards finding a cube root, but they assume that your answer will be expressed as a whole number.
Link 1: https://www.math.hmc.edu/~benjamin/papers/Cubing.pdf
Link 2: http://www.careeranna.com/calculate-cube-roots-in-less-than-5-seconds/
