Did Feynman mentally compute $\sqrt[3]{1729.03}$ by linear approximation?

Feynman tells the story in one of his books of anecdotes.

http://www.ee.ryerson.ca/~elf/abacus/feynman.html

$12$ is a very good first approximation and the linear term of the series expansion suffices to get high precision.

$$ \sqrt[3]{1728 + d} = 12\sqrt[3]{1+x} = 12 + 4x + O(x^2)$$

where $d = 1.03$ and $x = \frac{d}{1728}$ is, in Feynman's words, about 1 part in 2000, so that the error term is of order $10^{-6}$.

Feynman says that he computed $12 + \frac{4d}{1728}$ as the approximate value.

The number was 1729.03. I happened to know that a cubic foot contains 1728 cubic inches, so the answer is a tiny bit more than 12. The excess, 1.03 is only one part in nearly 2000, and I had learned in calculus that for small fractions, the cube root's excess is one-third of the number's excess. So all I had to do is find the fraction 1/1728, and multiply by 4 (divide by 3 and multiply by 12). So I was able to pull out a whole lot of digits that way.

He describes that as though $d=1$ for this part of the calculation, so maybe $12 + \frac{1}{432}$ was what he actually computed. By "adding two more digits" (to 12.002) he seems to mean working out the division in the fraction. It could also mean adding (0.03)/432 as two more decimal digits of accuracy to $(12 + 432^{-1})$, which requires only a multiplication by 3 of an already computed quantity 1/432.

Feynman's method is the one that would have been immediate for anyone familiar with the binomial series and with $12^3 = 1728$. He said that knew the latter as ft^3/in^3 and other people might know it from the Ramanujan 1729 story. The other ingredient, as Feynman says in the story, was being good at integer division.

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 $\dfrac{.34332}{12^2}=\dfrac{.05722}{12*2}=\dfrac{.02861}{12}\approx.002384$.