204. It frequently occurs that the terms of the binomial are affected with co-efficients and exponents, as in the following example:

Let it be required to raise the binomial

(x + y) = x2 + 4x3y + 6x2y2 + 4xy3 + y2;

and substituting for x and y their values, we have

(3a2c2bd)+= (3a2c) + 4 (3a2c)3 (− 2bd) + 6 (3a2c)2 (— 2bd)2 +4(3a2c) (-2bd)3 + (−2bd)*,

or, by performing the operations indicated,

(3a2c - 2bd)1 = 81ac4 — 216a6c3bd + 216a1c2b2d2 — 96a2cb3d3

The terms of the development are alternately plus and minus, as they should be, since the second term is

205. The powers of any polynomial, may easily be found by the binomial theorem. For example, raise

to the third power.

First, put

Then

a+b+c

b + c = d.

(a+b+c)3 = (a + d)3 = a3 + 3a2d + 3ad2 + d3;

and by substituting for the value of d,

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

3a2c +362c+6abc

+3ac23bc2
+ c3.

This development is composed of the cubes of the three terms, plus three times the square of each term by the first powers of the two others, plus six times the product of all three terms.

To apply the preceding formula to the development of the cube of a trinomial, in which the terms are affected with co-efficients and exponents, designate each term by a single letter, and perform the operations indicated; then replace the letters introduced by their values

The fourth, fifth, &c. powers of any polynomial can be developed in a similar manner.

Consequences of the Binomial Formula.

206. The development of the binomial expression (x + a)" will always contain m +1 terms. Hence, if we take that term of the development which has n terms before it, the number of terms after it will be expressed by m — n.

Let us now seek the co-efficient of the term which has n terms after it, and which, consequently, has m n terms before it. We obtain this co-efficient by simply substituting m in the last value of N in Art. 203.

We then have,

(n + 2) (n + 1)

1 2 3 (m n

1) (m — n)'

n for n,

As we can always take the term which has n terms before it, nearer to the first term than the one which has m n terms before it, we will examine that part of the co-efficient which is derived from the terms lying between these two. We may write

N=

m(m—1) . . . (m—n+1). (m—n). (m—n−1)... (n+2). (n+1)
1. 2
(n+1). (n+2).. (m-n-1). (m-n)'

n

Now, by cancelling the like factors in the numerator and denominator, we have

In the development of any power of a binomial, the co-efficients at equal distances from the two extremes are equal to each other.

207. If we designate by K the co-efficient of the term which has n terms before it, that term will be expressed by Kanam-"; and the corresponding term counted from the last term of the series, will be Kam-non.

Now, the first co-efficient expresses the number of different combinations that can be formed with m letters taken n and n; and the second, the number which can be formed when taken m

--

and m

-

n.

and 5, give the same

n; we may therefore conclude that, the number of dif ferent combinations of m letters taken n and n, is equal to the number of combinations of m letters taken m n and m For example, twelve letters combined 5 number of combinations as when taken 12 5 and 12 5, or 7 and 7. Five letters combined 2 and 2, give the same number of combinations as when combined 5 2 and 52, or 3 and 3.

208. If, in the general formula,

(x + a)m = xm + maxm−1 + m

That is, the sum of the co-efficients of all the terms of the formula for the binomial, is equal to the mth power of 2.

Thus, in 'the particular case

(x + a)5 = x2 + 5ax1 + 10a2x3 + 10a3x2 + 5a1x + a3, the sum of the co-efficients

1 + 5 + 10 + 10 + 5 + 1

is equal to 25 = 32. In the 10th power developed, the sum of the co-efficients is equal to 210 — 1024.

Extraction of the Cube Root of Numbers.

209. The cube or third power of a number, is the product which arises from multiplying the number twice by itself. The cube root, or third root of a number is either of three equal factors into which it may be resolved; and hence, to extract the cube root, is to seek one of these factors.

Every number which can be resolved into three equal factors that are commensurable with unity, is called a perfect cube; and any number which cannot be so resolved, is called an imperfect cube.

The first ten numbers are

roots, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10;

cubes,

1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Reciprocally, the numbers of the first line are the cube roots of the numbers of the second

We perceive, by inspection, that there are but nine perfect cubes among all the numbers expressed by one, two, and three figures. Every other number, except the nine written above, which can be expressed by one, two, or three figures, will be an imperfect cube; and hence, its cube root will be expressed by a whole number, plus an irrational number, as may be shown by a course of reasoning entirely similar to that pursued in the latter part of Art. 118. 210. Let us find the difference between the cubes of two consecutive numbers.

Let a and a + 1, be two consecutive whole numbers; we have (a + 1)3 = a3 + 3a2 + 3a + 1;

whence,

(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 the number, plus 1.

Thus, the difference between the cube of 90 and the cube of 89, is equal to

3 (89)2 + 3 × 89 + 1 = 24031.

211. In order to extract the cube root of an entire number, we will observe, that when the figures expressing the number do not exceed three, the entire part of the root is found by comparing the number with the first nine perfect cubes. For example, the cube root of 27 is 3. The cube root of 30 is 3, plus an irrational number, less than unity. The cube root of 72 is 4, plus an irrational number less than unity, since 72 lies between the perfect cubes 64 and 125.

When the number is expressed by more than three figures, the process will be as follows. Let the proposed number be 103823.

This number being comprised between 1,000, which is the cube

of 10, and 1,000,000, which is the cube of 100, its root will be expressed by two figures, or by tens and units. Denoting the tens by a, and the units by b, we have (Art. 198),

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

Whence it follows, that the cube of a number composed of tens and units, is made up of four distinct parts: viz., the cube of the tens, three times the product of the square of the tens by the units, three times the product of the tens by the square of the units, and the cube of the units.

Now, the cube of the tens, giving at least, thousands, the last three figures to the right cannot form a part of it: the cube of tens must therefore be found in the part 103 which is separated from the last three figures. The root of the greatest cube contained in 103 being 4, this is the number of tens in the required root. Indeed, 103823 is evidently comprised between (40)3 or 64,000, 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 103, and there remains 39, to which bring down the part 823, and we have 39823, which contains three times the square of the tens by the units, plus the two parts named above. Now, as the square of tens gives at least hundreds, it follows that three times the square 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 398 by 48, which is three times the square of the tens, the quotient 8 will be the units 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 from the 4 tens and 8 units the three parts which enter into 39823; 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 its cube root.

REMARK I.-The units figures could not be first obtained, 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.