Manual Math: Long-Division Method for \(n\)th Roots
\[ \definecolor{vscolor}{RGB}{192,128,64} \definecolor{hlacolor}{RGB}{192,255,192} \definecolor{hlbcolor}{RGB}{192,192,255} \definecolor{hlccolor}{RGB}{128,255,255} \definecolor{hlPcolor}{RGB}{64,255,64} \newcommand{\hp}[1]{\hphantom{#1}} \newcommand{\vs}[]{ {\color{vscolor}┆} } \newcommand{\hla}[1]{ {\color{hlacolor}#1} } \newcommand{\hlb}[1]{ {\color{hlbcolor}#1} } \newcommand{\hlc}[1]{ {\color{hlccolor}#1} } \newcommand{\hlP}[1]{ {\color{hlPcolor}#1} } \newcommand{\eol}[]{\hspace{-0.04em}} \newcommand{\eolB}[]{\hspace{-0.02em}} \]Let's say you don't have a calculator and you want to extract the \(n\)th root[1] of a number. Chances are excellent you didn't learn how to do this in school[2], so it seems all hope is lost.
However, there is actually an arcane algorithm for computing \(n\)th roots digit-by-digit, in a manner similar to long-division[3]. Just to set expectations, it is most practical for square-roots, but unlike anywhere else on the internet, I will also explain how to do cube-roots (and beyond!).
Table of contents / quick links:
Square-Roots
The algorithm for square-roots is the easiest to understand, and the most practical. Unfortunately, it is still a bit convoluted, and is most clearly explained while working an example[4].
Simple Example: \(\sqrt{576}\)
Consider the problem:
\[ \sqrt{576} = \text{?} \]First, starting from the decimal point and proceeding both to the left and right, we group the digits by twos (intuitively, because this is the 2nd root):
\begin{align*} &\hspace{1.02em}\hp{0}\vs\hp{00}\vs\\[-0.5em] &\sqrt{5 \vs 76 \vs} \end{align*}We start at the leftmost group (which here is just \(\hlc{5}\)) as our working value, and find the largest one-digit square that goes into it. This is the first digit of our answer! In our example here, the largest square that fits is \(\hla{2}^2 = \hlb{4}\). Therefore, the value we squared, \(\hla{2}\), is the first digit! We also subtract off just like in long-division, except here we subtract off the square, \(\hlb{4}\):
\begin{align*} &\hspace{1.02em}\hla{2}\vs\hp{00}\vs\\[-0.5em] &\sqrt{\hlc{5} \vs 76 \vs}\\[-0.25em] \underline{\hla{2}} \times \underline{\hla{2}} = \hlb{4} \hspace{2em} &\hspace{0.28em}\underline{-\hlb{4}\eol}\vs\\[-0.25em] &\hspace{0.24em}\hp{-}1\vs \end{align*}Just like in long-division, we bring down the next group of digits. Note that because this is a square-root, there are two digits per group, so we bring down two digits at once:
\begin{align*} &\hspace{1.02em}\hla{2}\vs\hp{00}\vs\\[-0.5em] &\sqrt{5 \vs 76 \vs}\\[-0.25em] \underline{2} \times \underline{2} = 4 \hspace{2em} &\hspace{0.28em}\underline{-4\eol}\vs\downarrow\hspace{-0.064em}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}\hlc{1}\vs\hlc{76}\vs \end{align*}The current answer is \(\hla{2}\) and the working value is \(\hlc{176}\).
Now comes the key tricky step. First, we take the current answer and multiply it by \(20\) (yes; it will always be \(20\)). Then we want to think about multiplying by an unknown digit \(d\), and adding its square, \(d^2\). We want to find the largest such digit that makes the result still not exceed the working value.
For our example, the answer so far is \(\hla{2}\) and the working value is \(\hlc{176}\). We seek the largest one-digit \(d\) such that:
\begin{align*} (20\!\cdot\!\hla{2})d + d^2 &\leq \hlc{176} \\ (40+d) \cdot d &\leq \hlc{176} \end{align*}However, for square-roots, there is a more intuitive way we can do the same thing! Instead, we take the current answer, double it, and add a blank to the right for a new ones-place[5]. Then we add a multiplication by the same blank. For our example:
\[ (4\underline{ \hp{0} } \times \underline{ \hp{0} }) \leq \hlc{176} \]We solve this by trial multiplication. Here are all possible guesses for \(\hla{d}\):
\begin{align*} 4\underline{\hla{0}} \times \underline{\hla{0}} &= \hlb{\hp{00}0} \\ 4\underline{\hla{1}} \times \underline{\hla{1}} &= \hlb{\hp{0}41} \\ 4\underline{\hla{2}} \times \underline{\hla{2}} &= \hlb{\hp{0}84} \\ 4\underline{\hla{3}} \times \underline{\hla{3}} &= \hlb{ 129} \\ 4\underline{\hla{4}} \times \underline{\hla{4}} &= \hlb{ 176} \\ 4\underline{\hla{5}} \times \underline{\hla{5}} &= \hlb{ 225} \\ 4\underline{\hla{6}} \times \underline{\hla{6}} &= \hlb{ 276} \\ 4\underline{\hla{7}} \times \underline{\hla{7}} &= \hlb{ 329} \\ 4\underline{\hla{8}} \times \underline{\hla{8}} &= \hlb{ 384} \\ 4\underline{\hla{9}} \times \underline{\hla{9}} &= \hlb{ 441} \end{align*}Remember, we want the highest one that is no larger than the working value \(\hlc{176}\), so we select \(d=\hla{4}\), which happens to match it exactly. This is the next digit of our answer. We write the digit and subtract off the multiplied value \(\hlb{176}\):
\begin{align*} &\hspace{1.02em}2\vs\hp{0}\hla{4}\vs\\[-0.5em] &\sqrt{5 \vs 76 \vs}\\[-0.25em] \underline{2} \times \underline{2} = 4 \hspace{2em} &\hspace{0.28em}\underline{-4\eol}\vs\hp{00}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}1\vs76\vs\\[-0.25em] 4\underline{\hla{4}} \times \underline{\hla{4}} = \hlb{176} \hspace{2em} &\hspace{0.24em}\underline{\hlb{-1}\vs\hlb{76}\eolB}\vs\\[-0.5em] &\hspace{0.24em}\hp{-0}\vs\hp{0}\hlc{0}\vs \end{align*}The current working value is \(\hlc{0}\), and all the digit groups we could bring down (the implicit \(00\) groups to the right of the decimal point) are zero, so we are done!
\[ \boxed{ \sqrt{576} = 24 } \]You might be able to notice that back when we computed the first digit with \(\hla{2}^2=\hlb{4}\), that was really the same procedure, just with the initial answer being \(0\) because we didn't have anything yet! Twice zero is zero, and adding a blank to the right gives the multiplication problem we solved:
\[ (0\underline{\hla{2}} \times \underline{\hla{2}}) \leq \hlc{5} \]Harder Example: \(\sqrt{97 812.3}\)
If the answer does not come out exactly, the algorithm will not terminate. This next example shows a larger value and decimals, but all the mechanisms from above are the same.
Consider:
\[ \sqrt{97 812.3} = \text{?} \]First, we group to the left and right of the decimal point by twos:
\begin{align*} &\hp{~~~~3}\vs\hp{00}\vs\hp{00.}\vs\hp{00}\vs\\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs} \end{align*}The largest square that goes into the first group is \(\hla{3}^2=\hlb{9}\), so that's our first digit, we subtract off the square, and bring down the next group:
\begin{align*} &\hspace{1.02em}\hla{3}\vs\hp{00}\vs\hp{00.}\vs\hp{00}\vs\\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs}\\[-0.25em] \underline{\hla{3}} \times \underline{\hla{3}} = \hlb{9} \hspace{2em} &\hspace{0.28em}\underline{-\hlb{9}\eol}\vs\downarrow\hspace{-0.064em}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}\hlc{0}\vs\hlc{78}\vs \end{align*}The working value is \(\hlc{78}\). Take the current answer (i.e. \(\hla{3}\)), double it (i.e. \(6\)), and add a blank to the right to form the multiplication problem:
\[ (6\underline{\hp{0}} \times \underline{\hp{0}}) \leq \hlc{78} \]The largest digit that fits is \(d=\hla{1}\):
\begin{align*} (6\underline{\hla{1}} \times \underline{\hla{1}}) &\leq \hlc{78}\\ \hlb{61} &\leq \hlc{78} \end{align*}Therefore, the digit \(\hla{1}\) is the next digit of our answer. We write it, subtract the multiplication result \(\hlb{61}\), and bring the next group of digits down:
\begin{align*} &\hspace{1.02em}3\vs\hp{0}\hla{1}\vs\hp{00.}\vs\hp{00}\vs\\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs}\\[-0.25em] \underline{3} \times \underline{3} = 9 \hspace{2em} &\hspace{0.28em}\underline{-9\eol}\vs\hp{00}\vs\hp{00.}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}0\vs78\vs\hp{00.}\vs\\[-0.25em] 6\underline{\hla{1}} \times \underline{\hla{1}} = \hlb{61} \hspace{2em} &\hspace{1.34em}\underline{-\hlb{61}\eol}\vs\downarrow\hspace{0.23em}\vs\\[-0.5em] &\hspace{2.08em}\hlc{17}\vs\hlc{12\hp{.}}\vs \end{align*}The current working value is \(\hlc{1 712}\). Take the current answer (i.e. \(\hla{31}\)), double it (i.e. \(62\)), and add a blank to the right to form the multiplication problem:
\[ (62\underline{\hp{0}} \times \underline{\hp{0}}) \leq \hlc{1 712} \]The largest digit that fits is \(d=\hla{2}\):
\begin{align*} (62\underline{\hla{2}} \times \underline{\hla{2}}) &\leq \hlc{1 712}\\ \hlb{1244} &\leq \hlc{1 712} \end{align*}So, the digit \(\hla{2}\) is the next digit of our answer. We write it, subtract the multiplication result \(\hlb{1244}\), and bring the next group of digits down:
\begin{align*} &\hspace{1.02em}3\vs\hp{0}1\vs\hp{0}\hla{2.}\vs\hp{00}\vs\\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs}\\[-0.25em] \underline{3} \times \underline{3} = 9 \hspace{2em} &\hspace{0.28em}\underline{-9\eol}\vs\hp{00}\vs\hp{00.}\vs\hp{00}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}0\vs78\vs\hp{00.}\vs\hp{00}\vs\\[-0.25em] 6\underline{1} \times \underline{1} = 61 \hspace{2em} &\hspace{1.34em}\underline{-61\eol}\vs\hp{00.}\vs\hp{00}\vs\\[-0.5em] &\hspace{2.08em}17\vs12\hp{.}\vs\hp{00}\vs\\[-0.25em] 62\underline{\hla{2}} \times \underline{\hla{2}} = \hlb{1 244} \hspace{2em} &\hspace{1.3em}\underline{-\hlb{12}\vs\hlb{44}\hp{.}}\vs\downarrow\hspace{-0.064em}\vs\\[-0.5em] &\hspace{2.08em}\hlc{\hp{0}4}\vs\hlc{68\hp{.}\eolB}\vs\hlc{30}\vs \end{align*}We continue the pattern, not stopping because this one doesn't work out exactly:
\begin{align*} &\hspace{1.02em}3\vs\hp{0}1\vs\hp{0}2.\vs\hp{0}7\vs\hp{0}4\vs\hp{0}9\vs\cdots\\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs00\vs00\vs\cdots}\\[-0.25em] 0\underline{3} \times \underline{3} = 9 \hspace{2em} &\hspace{0.28em}\underline{-9\eol}\vs\hp{00}\vs\hp{00.}\vs\hp{00}\vs\hp{00}\vs\hp{00}\vs\\[-0.5em] &\hspace{0.24em}\hp{-}0\vs78\vs\hp{00.}\vs\hp{00}\vs\hp{00}\vs\hp{00}\vs\\[-0.25em] 6\underline{1} \times \underline{1} = 61 \hspace{2em} &\hspace{1.34em}\underline{-61\eol}\vs\hp{00.}\vs\hp{00}\vs\hp{00}\vs\hp{00}\vs\\[-0.5em] &\hspace{2.08em}17\vs12\hp{.}\vs\hp{00}\vs\hp{00}\vs\hp{00}\vs\\[-0.25em] 62\underline{2} \times \underline{2} = 1 244 \hspace{2em} &\hspace{1.3em}\underline{-12\vs44\hp{.}\eolB}\vs\hp{00}\vs\hp{00}\vs\hp{00}\vs\\[-0.5em] &\hspace{2.08em}\hp{0}4\vs68\hp{.}\vs30\vs\hp{00}\vs\hp{00}\vs\\[-0.25em] 6 24\underline{7} \times \underline{7} = 43 729 \hspace{2em} &\hspace{1.8em}\underline{-4\vs37\hp{.}\vs29\eol}\vs\hp{00}\vs\hp{00}\vs\\[-0.5em] &\hspace{2.08em}\hp{00}\vs31\hp{.}\vs01\vs00\vs\hp{00}\vs\\[-0.25em] 62 54\underline{4} \times \underline{4} = 250 176 \hspace{2em} &\hspace{2.86em}\underline{-25\hp{.}\vs01\vs76\eol}\vs\hp{00}\vs\\[-0.5em] &\hspace{4.14em}5\hp{.}\vs99\vs24\vs00\vs\\[-0.25em] 625 48\underline{9} \times \underline{9} = 5 629 401 \hspace{2em} &\hspace{3.36em}\underline{-5\hp{.}\vs62\vs94\vs01\hspace{-0.06em}}\vs\\[-0.5em] &\hspace{5.48em}36\vs29\vs99\vs\\[-0.25em] \vdots \hspace{6.3em} & \hspace{9em}\vdots \end{align*} \[ \boxed{ \sqrt{97 812.3} = 312.749\cdots } \]Cube-Roots
The procedure for taking cube-roots works similarly. Because we are taking the 3rd root instead of the 2nd root, we group the digits by threes instead of by twos. Additionally, the procedure for finding the unknown digits is more complicated.
For square-roots, the procedure at step \(j\) was (where \(\hla{A_j}\) is the current answer, \(\hlc{V_j}\) is the working value, and \(d\) is the unknown next answer digit[6]) was to solve the following polynomial inequality for the largest digit \(d\):
\[ p_2(d) \leq \hlc{V_j} \text{, where } p_2(d) = (20\hla{A_j})d + d^2 \hspace{6em} \]For cube-roots, the polynomial is:
\[ p_3(d) \leq \hlc{V_j} \text{, where } p_3(d) = (300\hla{A_j}^2) d + (30\hla{A_j})d^2 + d^3 \]For square-roots, solving the polynomial simplifies to the blanks thing, but for cube-roots, there appears to be no analogously intuitive simplification. For trial multiplication, I recommend working with the first term first, since it is the largest and most significant.
Example: \(\sqrt[3]{75 686 967}\)
Consider the problem:
\[ \sqrt[3]{75 686 967} = \text{?} \]We group by threes instead of twos:
\begin{align*} &\hspace{1.52em}\hp{0}\vs\hp{000}\vs\hp{000}\vs\\[-0.5em] &\sqrt[3]{75 \vs 686 \vs 967 \vs} \end{align*}Then, we find the largest cube that fits into the first group. As before, notice that this is just a special case of the general procedure for solving the polynomial. We have no answer digits yet, so those terms are zero:
\begin{align*} \cancel{(300 \cdot 0^2 \cdot d)} + \cancel{(30 \cdot 0 \cdot d^2)} + d^3 &\leq \hlc{75} \\ d^3 &\leq \hlc{75} \end{align*}We find that the largest digit is \(d=\hla{4}\). That is our first digit. We subtract off the result of the procedure and bring down the next group:
\begin{align*} &\hspace{1.52em}\hla{4}\vs\hp{000}\vs\hp{000}\vs\\[-0.5em] &\sqrt[3]{75 \vs 686 \vs 967 \vs}\\[-0.25em] (300 \cdot 0^2 \cdot \underline{\hla{4}}) + (30 \cdot 0 \cdot \underline{\hla{4}}^2) + \underline{\hla{4}}^3 = \hlb{64} \hspace{2em} &\hspace{0.28em}\underline{-\hlb{64}\eol}\vs\hspace{0.22em}\downarrow\hspace{0.22em}\vs\\[-0.5em] &\hspace{1.02em}\hlc{11}\vs\hlc{686}\vs \end{align*}The next instance of the procedure, we do have digit(s)—namely the \(4\) we just found:
\begin{align*} (300 \cdot 4^2 \cdot d) + (30 \cdot 4 \cdot d^2) + d^3 &\leq \hlc{11 686} \\ 4800 d + 120 d^2 + d^3 &\leq \hlc{11 686} \end{align*}Just looking at the first term, we see that \(d\) can't be bigger than \(2\), and indeed this is the correct largest value:
\begin{align*} &\hspace{1.52em}4\vs\hp{00}\hla{2}\vs\hp{000}\vs\\[-0.5em] &\sqrt[3]{75 \vs 686 \vs 967 \vs}\\[-0.25em] (300 \cdot 0^2 \cdot \underline{4}) + (30 \cdot 0 \cdot \underline{4}^2) + \underline{4}^3 = 64 \hspace{2em} &\hspace{0.28em}\underline{-64\eol}\vs\hp{000}\vs\hp{000}\vs\\[-0.5em] &\hspace{1.02em}11\vs686\vs\hp{000}\vs\\[-0.25em] (300 \cdot 4^2 \cdot \underline{\hla{2}}) + (30 \cdot 4 \cdot \underline{\hla{2}}^2) + \underline{\hla{2}}^3 = \hlb{10 088} \hspace{2em} &\hspace{0.24em}\underline{-\hlb{10}\vs\hlb{088}\eolB}\vs\hspace{0.22em}\downarrow\hspace{0.22em}\vs\\[-0.5em] &\hspace{1.52em}\hlc{1}\vs\hlc{598}\vs\hlc{967}\vs \end{align*}One more time:
\begin{align*} &\hspace{1.52em}4\vs\hp{00}2\vs\hp{00}\hla{3}\vs\\[-0.5em] &\sqrt[3]{75 \vs 686 \vs 967 \vs}\\[-0.25em] (300 \cdot 0^2 \cdot \underline{4}) + (30 \cdot 0 \cdot \underline{4}^2) + \underline{4}^3 = 64 \hspace{2em} &\hspace{0.28em}\underline{-64\eol}\vs\hp{000}\vs\hp{000}\vs\\[-0.5em] &\hspace{1.02em}11\vs686\vs\hp{000}\vs\\[-0.25em] (300 \cdot 4^2 \cdot \underline{2}) + (30 \cdot 4 \cdot \underline{2}^2) + \underline{2}^3 = 10 088 \hspace{2em} &\hspace{0.24em}\underline{-10\vs088\eolB}\vs\hp{000}\vs\\[-0.5em] &\hspace{1.52em}1\vs598\vs967\vs\\[-0.25em] (300 \cdot 42^2 \cdot \underline{\hla{3}}) + (30 \cdot 42 \cdot \underline{\hla{3}}^2) + \underline{\hla{3}}^3 = \hlb{1 598 967} \hspace{2em} &\hspace{0.74em}\underline{-\hlb{1}\vs\hlb{598}\vs\hlb{967}\eol}\vs\\[-0.5em] &\hspace{2.01em}\vs\hp{000}\vs\hp{00}\hlc{0}\vs \end{align*} \[ \boxed{ \sqrt[3]{75 686 967} = 423 } \]Mathematical Details and Higher Roots
So far, we have the procedures for square-roots and cube-roots (where \(\hla{A_j}\) is the current answer, \(\hlc{V_j}\) is the working value, and \(d\) is the unknown next answer digit):
\[ p_n(d) \leq \hlc{V_j} \] \begin{alignat*}{3} p_2(d) &= (20\hla{A_j})d + d^2 \hspace{1em} && \text{(square-roots)} \\ p_3(d) &= (300\hla{A_j}^2) d + (30\hla{A_j})d^2 + d^3 \hspace{1em} && \text{(cube-roots)} \end{alignat*}If you're a curious sort, you might wonder what the procedure looks like for 4th, 5th, and higher roots. As a bit of a teaser, here are a few more, and I'll also factor and rewrite them a bit[7]:
\begin{align*} p_1(d) &= \hlP{1}(10A_j)^0 d^1 \\ p_2(d) &= \hlP{2}(10A_j)^1 d^1 + ~~\hlP{1}(10A_j)^0 d^2 \\ p_3(d) &= \hlP{3}(10A_j)^2 d^1 + ~~\hlP{3}(10A_j)^1 d^2 + ~~\hlP{1}(10A_j)^0 d^3 \\ p_4(d) &= \hlP{4}(10A_j)^3 d^1 + ~~\hlP{6}(10A_j)^2 d^2 + ~~\hlP{4}(10A_j)^1 d^3 + \hlP{1}(10A_j)^0 d^4 \\ p_5(d) &= \hlP{5}(10A_j)^4 d^1 + \hlP{10}(10A_j)^3 d^2 + \hlP{10}(10A_j)^2 d^3 + \hlP{5}(10A_j)^1 d^4 + \hlP{1}(10A_j)^0 d^5 \end{align*}. . . The coefficients on these are rows from Pascal's triangle (less the leading 1)!
In fact, this is not a coincidence, and comes from the way this algorithm can be derived. The particular mathematical procedure here is apparently due to a François Viète, around 1600[8][9].
We are trying to find the answer \(A\) which is the \(n\)th root of some radicand \(C\).
We proceed in steps. At step \(j\), we are looking at some prefix of the radicand \(C_j\) followed by the next group of digits \(N_j\). We also have our current answer / approximation \(A_j\) (which should be an underestimate, in the \(\leq\) sense), and an unaccounted-for remainder / error \(R_j = C_j - A_j^n\), which combines with the next group of digits to give \(V_j\) (used in the examples as the 'working value'). That's a lot of notation, so let's look at the penultimate step of the square-root above as an example:
\begin{align*} &\hspace{1.02em}\hla{3}\vs\hp{0}\hla{1}\vs\hp{00.}\vs\hp{00}\vs \hspace{2em} A_j = \hla{31} \\[-0.5em] &\sqrt{9\vs78\vs12.\vs30\vs} \hspace{2em} C_j=978,~~~ N_j=12 \\[-0.25em] &\hspace{0.28em}\underline{-9\eol}\vs\hp{00}\vs\hp{00.}\vs\hp{00}\vs \\[-0.5em] &\hspace{0.24em}\hp{-}0\vs78\vs\hp{00.}\vs\hp{00}\vs \\[-0.25em] &\hspace{1.34em}\underline{-\hlb{61}\eol}\vs\downarrow\hspace{0.23em}\vs\hp{00}\vs \\[-0.5em] &\hspace{2.08em}\hlc{17}\vs\hlc{12\hp{.}}\vs\hp{00}\vs \hspace{2em} R_j = \hlc{17},~~~ V_j=\hlc{1 712} \end{align*}How do we proceed to the next step? Well, first, we add a new group of digits to consideration. In principle, our tasks then are to expand the radicand, find a new digit \(d\) to expand our answer, and calculate the new remainder:
\begin{alignat*}{3} C_{j+1} &= 10^n C_j + N_j \hspace{2em} &&\text{(expanded radicand)} \\ A_{j+1} &= 10^n A_j + d \hspace{2em} &&\text{(expanded answer with new digit)} \\ R_{j+1} &= C_{j+1} - A_{j+1}^n \hspace{2em} &&\text{(updated remainder)} \end{alignat*}Expanding the radicand is not really anything: we just allow ourselves to consider another group of digits. As for the updated answer \(A_{j+1}\), recall that the game is to pick the new digit \(d\) so that the new remainder \(R_{j+1}\) is as small as possible (but never negative, because then it wouldn't be an underestimate anymore).
However, as written, it seems like computing \(R_{j+1}\) means we will have to compute \(A_{j+1}^n\)—and because we don't know the \(d\) that goes into the \(A_{j+1}\) in the first place, we have to compute this power for every \(d\) we need to try!
That is a huge amount of work! So instead, we compute the new remainder \(R_{j+1}\) by updating the old remainder \(R_j\):
\begin{align*} R_{j+1} &= C_{j+1} - A_{j+1}^n \\ &= C_{j+1} - A_{j+1}^n + 10^nR_j - 10^nR_j \\ &= (10^n C_j + N_j) - (10A_j + d)^n + 10^nR_j - 10^n(C_j - A_j^n) \\ &= N_j - (10A_j + d)^n + 10^nR_j + 10^n A_j^n \\ &= \left[~ 10^nR_j + N_j ~\right] + \left[~ (10A_j)^n - (10A_j + d)^n ~\right] \\ &= V_j + \left[~ (10A_j)^n - (10A_j + d)^n ~\right] \end{align*}At this point, we recall the binomial formula—
\[ (\alpha+\beta)^n = \sum_{k=0}^n \binom{n}{k} \alpha^{n-k} \beta^{k} \]—and use it to continue, expanding inside with \(\alpha:=10A_j\) and \(\beta:=d\):
\begin{align*} R_{j+1} &= V_j + \left[ (10A_j)^n - \sum_{k=0}^n \binom{n}{k} (10A_j)^{n-k} d^{k} \right] \\ &= V_j - \sum_{k=1}^n \binom{n}{k} (10A_j)^{n-k} d^{k} \\ \end{align*}If you squint at this and recall the examples, you should see that this is exactly what we are doing at each step! We bring down the next group of digits (i.e. \(+N_j\), which produces \(V_j=10^nR_j + N_j\)), and then we subtract off a polynomial. You should see that the polynomial is the same as the Pascal's triangle row thing above. (Note the binomial coefficient \(\binom{n}{k}\) is the \(k\)th entry of the \(n\)th row in Pascal's triangle.)
In particular, for \(n=2\) we have:
\begin{align*} R_{j+1} &= V_j - \sum_{k=1}^2 \binom{2}{k} (10A_j)^{2-k} d^{k} \\ &= V_j - \left[~ \binom{2}{1} (10A_j)^{2-1} d^{1} + \binom{2}{2} (10A_j)^{2-2} d^{2} ~\right] \\ &= V_j - \left[~ (2)(10) A_j d + (1)(1) d^2 ~\right] \\ &= V_j - \left[~ 20 A_j d + d^2 ~\right] \\ &= V_j - p_2(d) \end{align*}And for \(n=3\) we have:
\begin{align*} R_{j+1} &= V_j - \sum_{k=1}^3 \binom{3}{k} (10A_j)^{3-k} d^{k} \\ &= V_j - \left[~ \binom{3}{1} (10A_j)^{3-1} d^{1} + \binom{3}{2} (10A_j)^{3-2} d^{2} + \binom{3}{3} (10A_j)^{3-3} d^{3} ~\right] \\ &= V_j - \left[~ (3)(10^2) A_j^2 d + (3)(10) A_j d^2 + (1)(1) d^3 ~\right] \\ &= V_j - \left[~ 300 A_j^2 d + 30 A_j d^2 + d^3 ~\right] \\ &= V_j - p_3(d) \end{align*}Although this formulation is quite pretty in my view, unfortunately it should be said that it's practical mainly only for square-roots. Cube-roots are already pushing it, and by the time you get to fourth-roots, you're in serious trouble. As the power goes up, the polynomials get more complicated. Meanwhile, the number of digits in the groups goes up, making all the arithmetic larger and worse.
To solve these sorts of problems by hand, it may be better to resort to approximation—the subject, perhaps, of a future article.