TITLE: Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation?
QUESTION [18 upvotes]: In the biopic "infinity" about Richard Feynman. (12:54)
He computes $\sqrt[3]{1729.03}$ by mental calculation.
I guess that he uses linear approximation.
That is, 
he observe that $1728=12^3$.
Let $f(x)=\sqrt[3]{x}$.
Then $f'(x)=\frac{1}{3\sqrt[3]{x^2}}$ and $f'(1728)=\frac{1}{3\sqrt[3]{1728^2}}=\frac{1}{3\cdot 12^2}$.
Therefore, 
$$\sqrt[3]{1729.03}=f(1729.03)\approx f(1728)+f'(1728)(1729.03-1728)=12+\frac{1.03}{3\cdot 12^2}=12.002384\overline{259}.$$
Question 1. 
If he used the linear approximation, 
how did he compute $\frac{1.03}{3\cdot 12^2}=0.002384\overline{259}$ by a mental calculation?
Question 2.
If he didn't use the linear approximation, 
what is another method he might have used?

REPLY [3 votes]: I upvoted the other answer as it comes from the other book of anecdotes, but one way to calculate this without Feynman's experience might be the following. 
You can start with the linear approximation, and then you have to calculate $(\dfrac{1.03}{3})/12^2$. $\dfrac{1.03}{3}\approx .34333$, but that doesn't lend itself to division by $12^2$, so you can change the last digit to get $\dfrac{1.03}{3}\approx .34332$ 
(You know this will be helpful by the divisibility test for $3$, and can note it's also divisible by $2$ and $4$ if you're thinking ahead.)
Then $\dfrac{.34332}{12^2}=\dfrac{.05722}{12*2}=\dfrac{.02861}{12}\approx.002384$.