Cube Roots of Compound Expressions.

Since $42875$ has two groups, the root will have two figures.

The first group, $42$, contains the cube of the tens of the root.

The greatest cube in $42$ is $27$, and the cube root of $27$ is $3$. Hence $3$ is the tens' figure of the root.

We subtract $27$ from $42$, and bring down the next group, $875$. Since $a$ is $3$ tens or $30$, $3a^2=3\times {30}^2$, or $2700$. This trial-divisor is contained $5$ times in $15875$. The trial-divisor is completed by adding $3ab+b^2$; that is, $450+25$, to the trial-divisor.

The same method will apply to numbers of more than two groups of figures, by considering in each case $a$, the part of the root already found, as so many tens with respect to the next figure of the root.

12pt Extract the cube root of $57512456$.

$$\begin{gathered} *3r*4l\&\&\&57\&512\&456\&(386 \\ \&\&a^3=\&27 \\ \mathrm{\backslash cline}4-53a^2=\&3\times {30}^2=\&\mathrm{\backslash multicolumn}1@r|2700\&30\&512 \\ 3ab=\&3\times (30\times 8)=\&\mathrm{\backslash multicolumn}1@r|720 \\ {b}^2=\&8^2=\&\mathrm{\backslash multicolumn}1@r|64 \\ \mathrm{\backslash cline}3-3\&\&\mathrm{\backslash multicolumn}1@r|3484\&27\&872 \\ \mathrm{\backslash cline}4-6\&\&\mathrm{\backslash multicolumn}1@r|\&\phantom{0}2\&640\&456 \\ 3a^2=\&3\times {380}^2=\&\mathrm{\backslash multicolumn}1@r|433200 \\ 3ab=\&3\times (380\times 6)=\&\mathrm{\backslash multicolumn}1@r|6840 \\ {b}^2=\&6^2=\&\mathrm{\backslash multicolumn}1@r|36 \\ \mathrm{\backslash cline}3-3\&\&\mathrm{\backslash multicolumn}1@r|440076\&\phantom{0}2\&640\&456 \\ \mathrm{\backslash cline}4-6 \end{gathered}$$

Extract the cube root of $187.149248$.

$$\begin{gathered} *3r@*4rl\&\&\&\mathrm{\backslash multicolumn}2r187\text{.}\&149\&248\&(5.72 \\ \&\&a^3=\&\mathrm{\backslash multicolumn}2r125 \\ \mathrm{\backslash cline}4-63a^2=\&3\times {50}^2=\&\mathrm{\backslash multicolumn}1@r|7500\&\&62\&149 \\ 3ab=\&3\times (50\times 7)=\&\mathrm{\backslash multicolumn}1@r|1050 \\ {b}^2=\&7^2=\&\mathrm{\backslash multicolumn}1@r|49 \\ \mathrm{\backslash cline}3-3\&\&\mathrm{\backslash multicolumn}1@r|8599\&\&60\&193 \\ \mathrm{\backslash cline}4-7\&\&\&\mathrm{\backslash multicolumn}1@r|\&1\&956\&248 \\ 3a^2=\&3\times {570}^2=\&9747\&\mathrm{\backslash multicolumn}1@r|00 \\ 3ab=\&3\times (570\times 2)=\&34\&\mathrm{\backslash multicolumn}1@r|20 \\ {b}^2=\&2^2=\&\&\mathrm{\backslash multicolumn}1@r|4 \\ \mathrm{\backslash cline}3-4\&\&9781\&\mathrm{\backslash multicolumn}1@r|24\&1\&956\&248 \\ \mathrm{\backslash cline}5-7 \end{gathered}$$

It will be seen from the groups of figures that the root will have one integral and two decimal places.

If the given number is not a perfect cube, ciphers may be annexed, and a value of the root may be found as near to the true value as we please.

Extract the cube root of $1250.6894$.

$$\begin{gathered} rrrcclll\&\&\&\&1\&\mathrm{\backslash multicolumn}3l\text{250.689,400,(10.77} \\ \&\&a^3\&=\&1 \\ \mathrm{\backslash cline}4-63a^2=\&\mathrm{\backslash PadTo}3\times (1070\times 7)3\times {10}^2=\&\mathrm{\backslash multicolumn}1@r|\phantom{0}\phantom{0}300\&\&\&250\&\phantom{999}\&\phantom{999} \end{gathered}$$ Since $300$ is not contained in $250$, the next figure of the root will be $0$. $$\begin{gathered} rrr@ccrll3a^2=\&3\times {100}^2=\&\mathrm{\backslash multicolumn}1@r|30000\&\&\phantom{0}\&250\&689 \\ 3ab=\&3\times (100\times 7)=\&\mathrm{\backslash multicolumn}1@r|2100 \\ {b}^2=\&7^2=\&\mathrm{\backslash multicolumn}1@r|49 \\ \mathrm{\backslash cline}3-3\&\&\mathrm{\backslash multicolumn}1@r|32149\&\&\&225\&043 \\ \mathrm{\backslash cline}4-8\&\&\&\mathrm{\backslash multicolumn}1@r|\&\&25\&646\&400 \\ 3a^2=\&3\times {1070}^2=\&34347\&\mathrm{\backslash multicolumn}1@r|00 \\ 3ab=\&3\times (1070\times 7)=\&224\&\mathrm{\backslash multicolumn}1@r|70 \\ {b}^2=\&7^2=\&\&\mathrm{\backslash multicolumn}1@r|49 \\ \mathrm{\backslash cline}3-4\&\&34572\&\mathrm{\backslash multicolumn}1@r|19\&\&24\&200\&533 \\ \mathrm{\backslash cline}5-8\&\&\&\&\&1\&445\&867 \end{gathered}$$

Notice that if $a$ denotes the first term, and $b$ the second term of the root, the first complete divisor is $3a^2+3ab+b^2,$ and the second trial-divisor is $3(a+b{)}^2$, that is, $3a^2+6ab+3b^2.$

This expression may be obtained by adding to the preceding complete divisor, $3a^2+3ab+b^2$, its second term and twice its third term. Thus: $$\begin{gathered} rr3a^2+3ab+\&b^2 \\ 3ab+\&2b^2 \\ \mathrm{\backslash cline}1-23a^2+6ab+\&3b^2 \end{gathered}$$

This method of obtaining trial-divisors is of great importance for shortening numerical work, as may be seen in the following example:

Ex. Extract the cube root of $5$ to five places of decimals.

$$\begin{gathered} rrr@r@rcrrl\&\&\&\&\&5\text{.}\&\mathrm{\backslash multicolumn}3l000(1.70997 \\ \&\&\&\mathrm{\backslash multicolumn}3r{a}^3=1 \\ \mathrm{\backslash cline}4-73a^2=\&3\times {10}^2=\&300\&\mathrm{\backslash multicolumn}1@r|\&\&4\&000 \\ 3ab=\&3(10\times 7)=\&210\&\mathrm{\backslash multicolumn}1@r| \\ {b}^2=\&7^2=\&49\&\mathrm{\backslash multicolumn}1@r| \\ \mathrm{\backslash cline}3-3\&\&559\&\mathrm{\backslash multicolumn}1@r|\mathrm{\backslash BB}\&\&3\&913 \\ \mathrm{\backslash cline}5-9\&\&259\&\&\&\mathrm{\backslash multicolumn}1@r|\&87\&000\&000 \\ \mathrm{\backslash cline}3-53a^2=\&3\times {1700}^2=\&867\&00\&00\&\mathrm{\backslash multicolumn}1@r| \\ 3ab=\&3(1700\times 9)=\&4\&59\&00\&\mathrm{\backslash multicolumn}1@r| \\ {b}^2=\&9^2=\&\&\&81\&\mathrm{\backslash multicolumn}1@r| \\ \mathrm{\backslash cline}3-5\&\&871\&59\&81\&\mathrm{\backslash multicolumn}1@r|\mathrm{\backslash BB}\&78\&443\&829 \\ \mathrm{\backslash cline}7-9\&\&4\&59\&81\&\mathrm{\backslash multicolumn}1@r|\&8\&556\&1710 \\ \mathrm{\backslash cline}3-53a^2=\&3\times {1709}^2=\&876\&20\&43\&\mathrm{\backslash multicolumn}1@r|\&7\&885\&8387 \\ \mathrm{\backslash cline}7-9\&\&\&\&\&\mathrm{\backslash multicolumn}1@r|\&\&670\&33230 \\ \&\&\&\&\&\mathrm{\backslash multicolumn}1@r|\&\&613\&34301 \\ \mathrm{\backslash cline}7-9 \end{gathered}$$

After the first two figures of the root are found, the next trial-divisor is obtained by bringing down $259$, the sum of the $210$ and $49$ obtained in completing the preceding divisor, then adding the three lines connected by the brace, and annexing two ciphers to the result.