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Ex. 2. Find the cube root of 102503232.

102503232(468

64 3 x 49

48

38503 3 x 4 x 6 62

36 5556

33336 36

5167232 3 x 462

6348 3 x 46 x 8 1104 82 =

64 645904

5167232

EXPLANATION.—The first two figures of the root are obtained as in Ex. 1. We then treat the number they form, viza, 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 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 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 a? 3 ab 72

If we now again write down 63 Sum = 3 a2 + 3 ab + 62 under this sum, and then add up

62 ) the last four lines, we get3a? + 6 ab + 36, or 3 (a + 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

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

= 147

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 =

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 42 48

23000 3 x 4 x 4

48 42

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 12

1 58887991

58887991 1

280500900 58901283

235605132

44895768 Ans. 4.43104.

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Ex. XI. Find the square roots of 1. 4x*y*z, 16 a‘y, ** + 2 a 2 + a. 2. 4.26 - 12 ay + 29 *ya - 30 c®ya + 25 y.

2n - 3

30 ascm +

3. 25 a* 30 a3 + 19 a 12 6 ab3 + 64. 4. 1 4x + 10 22 20 203 + 25 2* 24 205 + 16 .

5. a? + 2 abx + (2 ac + 69) + 2 (ad + bc) 23 + (2 bd + c) ** + 2 cdac + d xc6.

6. a-xen 6 a®æan — 1 + 17 a 2n - 2 24 axen + 16 w21-4 7. 2 + 2 + x, aʻx – 2 2 + a 2xco.

a 8. 9 w2m - 3 a?cmm + 25 am

+ 5a.

4 Find the square roots of—

9. 1296, 6241, 42849, 83521. 10. 10650-24, 000576, 1, 48

2 13 + 1
1

1
Give the values correct to four places of decimals of—

of 31416 15 + 12 J10 - 2 12.

3.1416 of 193

√5 12 110 + 2

11. 117, V1.5, 15 + 1'73 – 1

40 32.16

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Find the cube roots of— 13. 8 a’l®y?, 125 xưay, a3 + 6 ab + 12 ab? + 86*. 14. 212 + 9 x10 + 6 208 99 206 42 2* + 441 x 343. 15. 208 + 3 x*y + 3 sya + ys

3 caca + 3 c^2c + 3 cʻy

16. 200 + 2-3 + 3 (x + x-2), ac*y+ 3 x*y-% + 3 xy-- +1. Find the cube roots of 17. 5849513501832, 1371.330631. 18. 20-346417; 037, 08

Give the value of the following correct to four places of decimals :

$5.12 + 9-03375 1 19.

980 - 9:01 34 + 32 + 1. 2/625 + 5.04 20.

of V:05 + 1.04

7 21. ( 17 + 2) ( 17 - 1), (5 + 13) (4 + 112).

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22. V11 + 6 72, 6 + 15 73. 3 6 + 2 15

15 + 1 23. 4

4 24. a(6 c) 33 (a – c) + c (a - b), where a = 6 V-3, and c = 9-027.

16

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CHAPTER IV.

GREATEST COMMON MEASURE AND LEAST COMMON MULTIPLE.

Greatest Common Measure.

44. In Arithmetic (page 24) we defined the G.C.M. of two or more numbers as their highest common factor.

In Algebra the same definition will suffice, provided we understand by the term highest common factor, the factor of highest dimensions (Art. 18). This, it need hardly be remarked, does not necessarily correspond to the factor of highest numerical value.

45. To find the G.C.M. of two quantities.

RULE.—Let A and B be the quantities, of which A is not of lower dimensions than B. Divide A by B, until a remainder is obtained of lower dimensions than B. Take this remainder as a new divisor, and the preceding divisor A as a new dividend, and divide till a remainder is again obtained of lower dimensions than the divisor; and so on. The last divisor is the G.C.M.

Before giving the general theory of the G.C.M. we shall work out a few examples.

Ex. 1. Find the G.C.M. of a 6x – 27 and 2c* - 11x – 63. According to the above rule, the operation is as follows:2 - 6. - 27)20 - 11 - 63(2 2 22 12 x – 54

27(3 + 3

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