Page:Elementary algebra (1896).djvu/200

1. Three times the square of $a$, the term of the root already found.

2. Three times the product of this first term a and the new term $b$.

3. The square of $b$.

The work may be arranged as follows:

$a^{3}+3a^{2}b+3ab^{2}+b^{3}(a+b$

$a^{3}$

$3(a^{2})=3a^{2}$ $3a^{2}b+3ab^{2}+b^{3}$

$3\times a\times b=+3ab$

$(b)^{2}=$ $+b^{2}$

$3a^{2}+3ab+b^{2}3a^{2}b+3ab^{2}-b^{2}$

Ex. 1. Find the cube root of $8x^{3}-36x^{2}y+54xy^{2}-27y^{3}$

$8x^{3}-36x^{2}y+54xy^{2}-27y^{3}(2x-3y)$

$8x^{3}$

$3(2x)^{2}=12x^{2}$

$3\times 2x\times (-3y)=-18xy$

$(-3y)^{2}=+9y^{2}$

$12x^{2}-18xy+9y^{2}$

$-36x^{2}y+54xy^{2}-27y^{3}$

$-36x^{y}+54xy^{2}/2-27y^{3}$

Explanation. The cube root of $8\times 3$ is $2x$, and this is the first term of the root.

By taking three times the square of this first term we obtain $12x^{2}$, which is the first term of the divisor, and is called the "trial divisor." Divide $-36x2y$, the first term of the remainder, by $12\times 2$ and we get $-3y$, the new term in the root. To complete the divisor, we first annex to the trial divisor three times the product of $2x$, the part of the root already found, and $-3y$, the new term of the root: this is $-18xy$. We then annex the square of $-3y$, the new term, and the divisor is complete. We next multiply this divisor by the new term, and subtract the result from the first remainder. There is now no remainder and the root has been found.

The process can be extended so as to find the cube root of any multinomial. The first two terms of the root will be obtained as before. When we have brought down the second remainder, we form the trial divisor by taking three times the square of the two terms of the root already found, and proceed as is shown in the following example.