Cube Roots of Compound Expressions.

Since the cube of $a+b$ is $a^3+3a^{2}b+3a{b}^2+b^3$, the cube root of a${}^3+3a^{2}b+3a{b}^2+b^3$ is $a+b$.

It is required to devise a method for extracting the cube root $a+b$ when $a^3+3a^{2}b+3a{b}^2+b^3$ is given.

1. Find the cube root of $a^3+3a^{2}b+3a{b}^2+b^3$.

$$\begin{gathered} rl<rcll\&\&a^3\&+\&\mathrm{\backslash multicolumn}1@r|3a^{2}b+3a{b}^2+b^3\&a+b \\ \mathrm{\backslash cline}6-63a^2\&\&a^3 \\ \mathrm{\backslash cline}3-3\&+3ab+b^2\&\mathrm{\backslash multicolumn}1@r|\&\&3a^{2}b+3a{b}^2+b^3 \\ \mathrm{\backslash cline}1-33a^2\&+3ab+b^2\&\mathrm{\backslash multicolumn}1@r|\&\&3a^{2}b+3a{b}^2+b^3 \\ \mathrm{\backslash cline}4-5 \end{gathered}$$

The first term $a$ of the root is obviously the cube root of the first term $a^3$ of the given expression.

If $a^3$ be subtracted, the remainder is $3a^{2}b+3a{b}^2+b^3$; therefore, the second term $b$ of the root is obtained by dividing the first term of this remainder by three times the square of $a$.

Also, since $3a^{2}b+3a{b}^2+b^3=(3a^2+3ab+b^2)b$, the complete divisor is obtained by adding $3ab+b^2$ to the trial divisor $3a^2$.

2. Find the cube root of $8x^3+36x^{2}y+54x{y}^2+27y^3$.

$$\begin{gathered} rlrlcll\&\&\&8x^3\&+\&\mathrm{\backslash multicolumn}1@r|36x^{2}y+54x{y}^2+27y^3\&2x+3y \\ \mathrm{\backslash cline}7-7\&12x^2\&\&8x^3 \\ \mathrm{\backslash cline}4-4(6x+3y)3y=\&\&18xy+\&\mathrm{\backslash multicolumn}1@r|9y^2\&\&36x^{2}y+54x{y}^2+27y^3 \\ \mathrm{\backslash cline}2-4\&12x^2\&+18xy+\&\mathrm{\backslash multicolumn}1@r|9y^2\&\&36x^{2}y+54x{y}^2+27y^3 \\ \mathrm{\backslash cline}5-6 \end{gathered}$$

The cube root of the first term is $2x$, and this is therefore the first term of the root. $8x^3$, the cube of $2x$, is subtracted.

The second term of the root, $3y$, is obtained by dividing $36x^{2}y$ by $3(2x{)}^2=12x^2$, which corresponds to $3a^2$ in the typical form, and the divisor is completed by annexing to $12x^2$ the expression $\backslash \{3(2x)+3y\backslash \}3y=18xy+9y^2,$ which corresponds to $3ab+b^2$ in the typical form.

The same method may be applied to longer expressions by considering $a$ in the typical form $3a^2+3ab+b^2$ to represent at each stage of the process the part of the root already found. Thus, if the part of the root already found is $x+y$, then $3a^2$ of the typical form will be represented by $3(x+y{)}^2$; and if the third term of the root be $+z$, then $3ab+b^2$ will be represented by $3(x+y)z+z^2$. So that the complete divisor, $3a^2+3ab+b^2$, will be represented by $3(x+y{)}^2+3(x+y)z+z^2$.

Ex. Find the cube root of $x^6-3x^5+5x^3-3x-1$.

$$\begin{gathered} \mathrm{\backslash ifthenelse}\text{⧸}\mathrm{\backslash boolean}ForPrinting*2r@*3r*4l\&\&\&\mathrm{\backslash multicolumn}1@r|\&x^2\&\mathrm{\backslash multicolumn}3@l-x-1 \\ \mathrm{\backslash cline}5-6\&\&\&\&x^6\&-3x^5\&\mathrm{\backslash multicolumn}2@l\phantom{+3x^4}+5x^3-3x-1 \\ \&3x^4\&\&\&x^6 \\ \mathrm{\backslash cline}5-5(3x^2-x)(-x)=\&\&-3x^3\&+\&\mathrm{\backslash multicolumn}1@r|x^2\&-3x^5\&\phantom{+3x^4}+5x^3 \\ \mathrm{\backslash cline}3-5\&3x^4\&-3x^3\&+\&\mathrm{\backslash multicolumn}1@r|x^2\&-3x^5\&+3x^4-\phantom{0}{x}^3 \\ \mathrm{\backslash cline}6-7\&\&\&\&\&\mathrm{\backslash multicolumn}1@r|\&-3x^4+6x^3-3x-1 \\ 3(x^2-x{)}^2=\&3x^4\&-6x^3\&+\&3x^2\&\mathrm{\backslash multicolumn}1@r| \\ \text{{} \end{gathered}$$(3x^{2} - 3x - 1)$$\begin{gathered} \}(-1)=\&\&\&-\&3x^2\&\mathrm{\backslash multicolumn}1l|+3x+1 \\ \mathrm{\backslash cline}2-6\&3x^4\&-6x^3\&\&\&\mathrm{\backslash multicolumn}1l|+3x+1\&-3x^4+6x^3-3x-1 \\ \mathrm{\backslash cline}7-7 \end{gathered}$$

The root is placed above the given expression because there is no room for it on the page at the right of the expression.

The first term of the root, $x^2$, is obtained by taking the cube root of the first term of the given expression; and the first trial-divisor, $3x^4$, is obtained by taking three times the square of this term.