For, from what has been said in No. 150. this expression represents the number of different combinations that can be formed of m letters taken p+1 and p+1. Now this number of combinations is, from its nature, an entire number; therefore the above expression is necessarily a whole number.

§ II. Of the extraction of the Roots of particular numbers.

156. The third power or cube of a number, is the product of this number multiplied by itself twice; and the third or cube root is the number which, raised to the third power, will produce the proposed number.

The first ten numbers being

their cubes are

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Reciprocally, the numbers of the second line have the numbers of the first line for their cube roots.

By inspecting these lines, we perceive that there are but nine perfect cubes among numbers composed of one, two, or three figures; each of the others has for its cube root a whole number, plus a fraction which cannot be expressed exactly by means

a

of unity. For suppose that the irreducible fraction can have

the whole number N for its root, it would follow that the cube of

аз

or would be equal to N. Now this is impossible, for

a and b being prime with respect to each other, a3 and b3 will

a 3

also be prime with respect to each other; therefore cannot

be equal to an entire number.

b3

157. The greater the roots of two consecutive perfect cubes are, the greater will be the difference between these cubes. Let a and a+1 be two consecutive whole numbers; we have (127)

whence

(a+1)3=a3+3a2+3a+1;

3

(a+1)3—a3 =3a2 +3a+1.

That is, the difference between the cubes of two consecutive

whole numbers is equal to three times the square of the least number, plus three times this number, plus 1.

Thus, the difference between the cube of 90 and the cube of 89, is equal to 3(89)2+3x 89+1=24031.

158. In order to extract the cube root of an entire number, we will observe, that when the number does not contain more than three figures, its root is obtained by merely inspecting the cubes of the nine first numbers. Thus, the cube root of 125 is 5; the cube root of 72 is 4 plus a fraction, or is within one of 4; the cube root of 841 is within one of 9, since 841 falls between 729, or the cube of 9, and 1000, or the cube of 10.

When the number contains more than three figures, the process will be as follows. Let the proposed number be

This number being comprised between 1000, which is the cube of 10, and 1000000, which is the cube of 100, its root must be composed of two figures, or of tens and units. Denoting the tens by a, and the units by b, we have (46)

3

(a+b)3=a3 +3a2b+3ab2 +b3.

Whence it follows, that the cube of a number composed of tens and units, contains the cube of the tens, three times the product of the square of the tens by the units, three times the product of the square of the units by the tens, plus the cul the units.

This being the case, the cube of the tens, giving at thousands, the three last figures to the right cannot for

part of it, the cube of the tens must therefore be found in the part 103 which is separated from the three last figures by a point. Now the root of the greatest cube contained in 103 being 4, this is the number of tens in the required root; for 103823 is evidently comprised between (40)3 or '64000, and (50)3 or 125,000; hence the required root is composed of 4 tens, plus a certain number of units less than ten.

Having found the number of tens, subtract its cube 64 from from 103; there remains 39, and bringing down the part 823, we have 39823, which contains three times the square of the tens by the units, plus the two parts before mentioned. Now as the square of a number of tens gives at least hundreds, it follows that three times the squares of the tens by the units, must be found in the part 398 to the left of 23, which is separated from it by a point. Therefore, dividing 348 by three times the square of the tens, which is 48, the quotient 8, will be the unit of the root, or something greater, since 398 hundreds is composed of three times the square of the tens by the units, together with the two other parts. We may ascertain whether the figure 8 is too great, by forming the three parts which enter in 39823, by means of the figure 8 and the number of tens 4; but it is much easier to cube 48, as has been done in the above table. Now the cube of 48 is 110592, which is greater than 103823; therefore 8 is too great. By cubing 47 we obtain 103823; hence the proposed number is a perfect cube, and 47 is the cube root of it.

N. B. The units figures could not be obtained first; because the cube of the units might give tens, and even hundreds, and the tens and hundreds would be confounded with those which arise from other parts of the cube.

Again, extract the cube root of 47954

The number 47954 being below 1000000, its root contains only two figures, viz. tens and units. The cube of the tens is found in 47 thousands, and we can prove, as in the preceding example, that 3, the root of the greatest cube contained in 47, expresses the tens. Subtracting the cube of 3 or 27 from 47 there remains 20; bringing down to the right of this remainder the figure 9 from the part 954, the number 209 hundreds is composed of three times the square of the tens by the units, plus the number arising from the other two parts. Therefore by forming three times the square of the tens, 3, which is 27, and dividing 209 by it, the quotient 7 will be the units of the root, or something greater. Cubing 37, we have 50653, which is greater than 47954; then cubing 36, we obtain 46656, which subtracted from 47954, gives 1298 for a remainder. Hence the proposed number is not a perfect cube; but 36 is its root to within unity. In fact, the difference between the proposed number and the cube of 36, is, as we have just seen, 1298, which is less than 3(36)2+3x36+1, for in verifying the result we have obtained 3888 for three times the square of 36.

159. Again, take for another example, the number, 43725658 containing more than 6 figures.

Whatever may be the required root, it contains more than one figure, and it may be considered as composed of units and tens only, (the tens being expressed by one or more figures.)

Now the cube of the tens gives at least thousands; it must therefore be found in the part which is to the left of the three

last figures 658. I say now that if we extract the root of the greatest cube contained in the part 43725, considered with reference to its absolute value, we will obtain the whole number of tens of the root; for let a be the root of 43725, to within unity, that is, such that 43725 shall be comprised between a3 and (a+1); then will 43725000 be comprehended between a3× 1000 and (a+1)3× 1000; and as these last two numbers differ from each other by more than 1000, it follows that the proposed number itself, 43725658, is comprised between a3×1000 and (a+1)3× 1000; therefore the required root is comprised between that of a3×1000, and (a+1)3×1000, that is, between a × 10 and (a+1)× 10. It is therefore composed of a tens, plus a certain number of units less than ten.

The question is then reduced to extracting the cube root of 43725; but this number having more than three figures, its root will contain more than one, that is, it will contain tens and units. To obtain the tens, point off the three last figures, 725, and extract the root of the greatest cube contained in 43.

The greatest cube contained in 43 is 27, the root of which is 3; this figure will then express the tens of the root of 43725, (or the figure in the place of hundreds in the total root). Subtracting the cube of 3, or 27, from 43, we obtain 16 for a remainder, to the right of which bring down the first figure 7, of the second period 725, which gives 167.

Taking three times the square of the tens, 3, which is 27, and dividing 166 by it, the quotient 6 is the unit figure of the root of 43725, or something greater. It is easily seen that this number is in fact too great; we must therefore try 5. The cube of 35 is 42875, which, subtracted from 43725, gives 850 for a remainder, which is evidently less than 3× (35)2 + 3×35 +1. Therefore 35 is the root of the greatest cube contained in 43725; hence it is the number of tens in the required root.

To obtain the units, bring down to the right of the remainder 850, the first figure, 6, of the last period, 658, which gives 8506; then take 3 times the square of the tens, 35, which is 3675, and divide 8506 by it; the quotient is 2, which we try by cubing 352: this gives 43614208, which is less than the proposed number, and subtracting it from this number, we obtain 111450 for a remainder. Therefore 352 is the cube root of 43725658, to within unity.