Consider the perfect cube 196122941. Its cube root is 581.
If I try to find out the cube root by the long division method, I do get 581 as follows:
$$ \begin{array}{r|r} & \underline{5 \quad 8 \quad 1} \quad\quad\quad\quad \quad \quad\quad \quad\quad\quad \quad \quad \quad\quad\\ \hline 5 & \overline{196}\quad\overline{122}\quad\overline{941} \quad \quad\quad \quad \quad\quad \quad\quad\quad \quad \quad&\\ &-125 \quad \quad\quad\quad \quad \quad\quad \quad\quad\quad \quad \quad\quad\quad \quad \quad\\ \\ \hline ((300\times5^2)+8=)7508 & 71 \quad 122 \quad \quad\quad\quad\quad \quad \quad\quad \quad\quad\quad \quad \quad\quad\\ & -70 \quad 112 (=(30(5)(58)+8^2)\times8)\quad\quad \quad\\ \\ \hline ((300\times 58^2)+1=)1009201 & 1010 \quad 941 \quad \quad\quad\quad\quad \quad \quad \quad\quad \quad\quad \quad\quad \\ & -1010 \quad 941 (=(30(58)(581)+1^2)\times1)\quad \\ \\ \hline & 0 \quad \quad \quad\quad \quad \quad\quad\quad \quad \quad\quad\\ \end{array} $$
where 7508 & 1009201 are estimated divisors from the formula $300a^2+b$. For 7508, $a$ is the 1st digit of quotient which is 5 & for 1009201, $a$ is the 1st 2 digits of quotient which is 58.The values for $b$ in each case is estimated such that $(30a(10a+b)+b^2)b$ is the highest positive integer $\le 71122$ & $\le 1010941$ in each case respectively. 70112 is thus obtained from the formula $(30a(10a+b)+b^2)b$ where $a=5,b=8$. 1009201 is obtained from the same latter formula but $a=58,b=1$.
I get the same result by the following division method too:
$$ \begin{array}{r|r|} & \underline{5 \quad 8 \quad 1}\quad \quad \quad \quad\quad \quad\quad\quad \quad\quad \quad \quad\quad \quad \quad\quad\quad \quad\quad \quad\\ \hline 5 & \overline{196}\quad\overline{122}\quad\overline{941} \quad \quad \quad \quad\quad \quad\quad\quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \\ & -125 \quad \quad\quad \quad \quad \quad \quad\quad \quad\quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\\ \\ \hline ([(5+5+5)\times10]+8=)158 & 71 \quad 122 \quad \quad \quad\quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\\ & -70 \quad 112 (=[(15\times8\times5)\times100]+[158\times8\times8])\quad \quad\\ \\ \hline ([(158+8+8)\times10 ]+1=)1741 & 1010 \quad 941 \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \\ & -1010 \quad 941 (=[(174\times1\times58)\times100]+[1741\times1\times1])\\ \\ \hline & 0 \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \\ \end{array} $$
where if we take a number on right side,say 158 ,it can be written as $A=A_{2}10^2+A_{1}10^1+A_{0}10^0$ where $A_{1}=1,A_{2}=5,A_{3}=8$. So any number on the right side can be thought as A=$A_{n}10^n...+A_{1}10^1+A_{0}10^0$.So it can thought as $A=10(B+2A_{0})+C$ where $B$ is the corresponding number in the adjacent upper square & $C$ is a number that is estimated. For instance here, $B$ for 158 is 5 while $B$ for 1741 is 158.
The value for $C$ in each case is estimated such that $100(A-A_{0}10^0)Ca+(A+A_{0}^2) $ is the highest positive integer $\le 71122$ & $\le 1010941$ in each case respectively. Here $a$ is the 1st digit of quotient which is 5 in the case of 158 & 1st 2 digits of the quotient which is 58 in the case of 1741. I have not used here the term 'estimated divisors' for 158 & 1741 because I seriously think it wouldn't be correct to do so.
What is the name of this second division method or what are the real names of these methods so as to distinguish between the 2 methods? Or are these methods basically the same in some sense?
