Cube root

Extracting cube roots by hand

Start by grouping the digits under the radical into groups of three as you normally would. This seems natural to do, even if it is a little more

${\sqrt[{3}]{10,\!460,\!353,\!203}}$

The first step here is to find the largest digit whose cube does not exceed the first group of digits under the radical, cube it, subtract to find the remainder, and bring down the next three digits.

${\begin{array}{rrr|rrrr}&&&2\\\hline &&{\sqrt[{3}]{}}&10&460&353&203\\{\underline {2}}^{3}&=&8&-8&\downarrow \\\hline &&&2&460\end{array}}$

The next step is based on the identity

$(10x+y)^{3}=1000x^{3}+300x^{2}y+30xy^{2}+y^{3}$

if we let $x=2$, the number written so far above the radical, and y be the next digit to calculate. The term $1000x^{3}=8000$ has already been subtracted when we bring down the next group of three digits, so the next digit to write above the radical will be the largest y such that $300\times 2^{2}y+30\times 2y^{2}+y^{3}\leq 2460$.

${\begin{array}{rrr|rrrr}&&&2&1\\\hline &&{\sqrt[{3}]{}}&10&460&353&203\\{\underline {2}}^{3}&=&8&-8&\downarrow \\\hline 300\times 2^{2}\times {\underline {1}}&&&2&460\\{}+30\times 2\times {\underline {1}}^{2}+{\underline {1}}^{3}&=&1261&-1&261&\downarrow \\\hline 300\times 21^{2}\times \_&&&1&199&353\\{}+30\times 21\times \_^{2}+\_^{3}&=&\end{array}}$

$${\begin{array}{rrr|rrrr}&&&2&1&8\\\hline &&{\sqrt[{3}]{}}&10&460&353&203\\{\underline {2}}^{3}&=&8&-8&\downarrow \\\hline 300\times 2^{2}\times {\underline {1}}&&&2&460\\{}+30\times 2\times {\underline {1}}^{2}+{\underline {1}}^{3}&=&1261&-1&261&\downarrow \\\hline 300\times 21^{2}\times 8&&&1&199&353\\{}+30\times 21\times 8^{2}+8^{3}&=&1099232&-1&099&232\\\hline &&&&100&121\end{array}}$$

To be continued ...