64 yo.

aa + 6x*y — 40x*y* + 96xy3 — 643o { ∞

x2 + 2xy

4y2

- 12x+y3- 48 x3y3 + 96 xy3 - 64 y

=

163*

3x2 + 12 x3y

−24xy3 + 16 y1

-12xy - 48 x3y+96xy-64 y

Ex. 3. Find the cube root of

2

It will be evident that the first two terms are obtained exactly as in Exs. 1 and 2. To
obtain the next term we treat the terms already found as corresponding to a in Ex. 1. Then
we obtain the trial divisor 3 (x2 + 2 xy)2 corresponding to 3 a2, and afterwards having divided
the first term of the remainder by the first term of the trial divisor in order to get the term
corresponding to b, we obtain in order 3 (x2 + 2 xy) (− 4 y2) corresponding to 3 ab, then (− 4 y2)2
corresponding to b2, and lastly, by addition, 3 x + 12 x3y 24 xy3 + 16 y, being the
complete divisor, corresponding to 3 a2 + 3 ab + b2 of Ex. 1. We then conclude the operation
as before.

Cube Root of Numerical Quantities.

41. We shall now apply the above method to numerical quantities. It may be shown, as in Art. 34, that, if we place a dot over the units' figure, and over every third figure to the left, the number of periods so formed will be the number of figures in the root.

EXPLANATION.-Pointing off the given number, we find the first period to be 262, and that the cube root consists of two figures. Now, the greatest perfect cube in 262 is 216, which is the cube of 6. Hence, the given number lies between the cubes of 60 and 70; and following the algebraical method, 60 will be the first term of the cube. This, we see, corresponds to a in the algebraical method.

We first then subtract the cube of a-viz., 216000, which leaves as a remainder 46144.

=

=

3 (60) (4)

=

720, and b2

We now write down 3 a2 or 3 (60) = 10800, which is the trial divisor for determining b. Dividing then by this value of 3 a2, we find b 4, which is the second term of the cube root. We next obtain 3 ab and so by addition we get 3 a2 + 3 ab + b2 is the complete divisor. Multiplying this then by the quotient figure, we subtract the product, and, there being no remainder, we find the cube root to be 64.

=

42 = 16, 11536, which

We may omit the useless ciphers in the above operation, if, remembering the local value of figures when numbers are expressed in ordinary notation, we take care to place the right-hand figure of the value of 3 ab one place to the right of the corresponding figure of the value of 3 a2; and also to place the right hand figure of b2 one figure further still to the right.

The operation will then stand thus

EXPLANATION.-The first two figures of the root are obtained as in Ex. 1. We then treat the number they form, viz., 46, as corresponding to a in the algebraical model, omitting useless ciphers. Obtaining then 3 a2 or 3 × 462 6348, we find 8 to be the next figure of the root. Then writing under this, 3 ab or 3 × 46 × 8 1104, and afterwards b2 or 82 = 64, taking care as to the positions of the right-hand figures, and adding, we get 645904 as the complete divisor. Then as before.

=

REMARK.—It is unnecessary to be at the trouble to find the value of 3 × 462 by ordinary multiplication. For referring to the algebraical model, and writing here the successive terms of the complete divisor, and adding, we have— 3 a2

3a2 + 6 ab + 36, or 3 (a2+2ab+b2) = 3 (a + b). This is three times the square of the first two terms of the root.

It therefore follows that, if, as in the above example, after completing the operation for finding the first two figures of the cube root, we write under the complete divisor just obtained the value of the square of the second figure, and then add together the last four lines thus obtained, we get three

times the square of the quotient for a partial divisor by which to determine the next figure of the root.

The four lines to be added are in the above example bracketed. This method will be found to materially shorten the work, for it may be similarly applied to find the trial divisor when the cube root consists of any number of figures

Cube Root of a Decimal.

42. We know that the cube of any number containing one, two, three, &c., decimal figures will contain three, six, nine, &c., decimal figures respectively, and hence, conversely, every decimal considered as a cube must contain a number of decimal figures which is a multiple of three, and the number of decimal figures in the cube root must be one-third of the number contained in the given cube. It will then be necessary to add ciphers when the given number of decimal figures is not a multiple of 3.

And by continuing the reasoning of Art. 37, if a dot be placed over the units' figure and over every third figure to the left, it will be sufficient to bring down the decimal figures three at a time, putting a decimal point in the quotient when the first three are brought down.

And further, if an integer be given which is not a perfect cube, we may proceed in the ordinary way till we arrive at a remainder, and then, putting a decimal point in the quotient, by affixing three ciphers to this and each successive remainder, approximate to the cube root as nearly as we please. Ex. 3. Find the cube root of 395-446904.

43. We shall show farther on that when n + 2 figures of a cube root have been obtained by the ordinary method, n figures more may be obtained by dividing the remainder by the next trial divisor, provided that the whole number of figures in the root is 2 n + 2.

We may apply this principle with advantage when we require the cube root of number to a given number of decimals. Ex. Find the cube root of 87 to five places of decimals. The required cube root will evidently contain 6 figures, and since 6 here corresponds to 2n + 2 above, it is evident that n 1=3 2. Hence, we shall find 4 (that is, n + 2) figures by the ordinary method, and then 2 more by division. The operation will stand thus

30 x3y3 + 25 x2y.