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Ex. 3. Find the cube root of —

26 + 6 xc" y - 40 2c3j3 + 96 xys

x*. - 64 y. **+ 6x*y 40x*y* + 96xy – 64 yo {** + 2xy

- 44 3** + 6 xy + 4x*y*

6."y

- 402*y

6'y + 12 y + 8x*y
3 (2c2 + 2xy)? 334 + 12xy + 122***

- 12 * y- 48 x*y8 + 96 xyó 64 76
3 (0? + 2xy) (- 4yo),

- 123*v* - 24.co
(-4y) =

16 y*
3.* + 12ary - 24 xy + 16 y* - 12 2 - 48 x + 96 2 64 vo
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) corresponding to 3 aʻ, and afterwards having divided
the first term of the remainder by the first term of the trial divisor in order to get the terin
corresponding to b, we obtain in order 3 (2+2 + 2 xy) (- 4 y) corresponding to 3 ab, then (- 4yo)
corresponding to b’, and lastly, by addition, 3 a* + 12 x*y - 24 xy + 16 y, being the
complete divisor, corresponding to 3 a* + 3 ab + b 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.

Ex. 1. Find the cube root of 262144.

262144(60 + $ = 64

216000
3 a = 10800

46144
3 ab 720
62

16
3 a2 + 3 ab + 62 11536 46144

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 seo, corresponds to a in the algebraical method.

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

We now write down 3 a2 or 3 (60)2 = 10800, which is the trial divisor for determining b. Dividing then by this value of 3 aʻ, we find b 4, which is the second term of the cube root. We next obtain 3 ab 3 (60) (4) = 720, and 6%

16, and so by addition we get 3 aa + 3 ab + 6 11536, which 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.

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 ao; and also to place the right hand figure of bone figure further still to the right. The operation will then stand thus

262144(64

216 3 x 62 108

46144 3 x 6 x 4 72 42

16 11536 46144

Ex. 2. Find the cube root of 102503232.

102503232(468

64 3 x 42

48

38503 3 x 4 x 6 72

62
5556

33336
36

5167232 3 x 462

6348 3 x 46 x 8 1104 82

64 645904

5167232

36

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 a’ or 3 x 46= 6348, we find 8 to be the next figure of the root. Then writing under this, 3 ab or 3 x 46 x 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 x 469 by ordinary multiplication. For referring to the algebraical model, and writing here the successive terms of the complete divisor, and adding, we have 3 a 3 ab 62

If we now again write down 12 Sum 3a + 3 ab + 12 ( under this sum, and then add up

the last four lines, we get3 a® + 6 ab + 369, or 3 (aż + 2 ab + b*) = 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.

395.446904(7.34

343 3 x 72 147

52446 3 x 7 x 3 63 32

9 15339

46017 9

6429904 3 x 732 15987 3 x 73 x 4

876 42

16 1607476 6429904

that n

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

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

87(4.43104

64 3 x 4? 48

23000 3 x 4 x 4

48

16 5296

21184 16

1816000 3 x 442 5808 3 x 44 x 3

396 32

9 584769

1754307 9

61693000 3 x 4432

588747 3 x 443 x 1

1329 1?

1 58887991

58887991 1

280500900 58901283

235605132

44895768 Ans. 4.43104.

Ex. XI. Find the square roots of— 1. 4x*yz, 16 a'y', * + 2 aʻa" + a'. 2. 4.26 - 12 y + 29 a*y* - 30 cy+ 25 x*y.

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