205. The powers of any polynomial may easily be found by the

binomial theorem.

For example, raise a+b+c to the third power.

This expression 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. It is easily proved that this law is true for any polynomial.

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, then replace the letters introduced, by their values, and perform the operations indicated. From this rule, we will find that

(2a2-4ab+36)3=8a-48a5b+132a*b-208a3b3
+198a2b-108ab5+27b.

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

Consequences of the Binomial Formula.

206. First. The expression (x+a)" being such, that a man be substituted for a, and a for 2, without altering its value, it follows that the same thing can be done in the development of it; therefore, if this development contains a term of the form Karam-n, it must have another equal to Karam-n or Ka"-"x". These two terms of the development are evidently at equal distances from the two extremes; for the number of terms which precede any term, being indicated by the exponent of a in that term, it follows that

the term Ka"xm-n has n terms before it; and that the term Kam-nx" has m-n terms before it, and consequently n terms after it, since the whole number of terms is denoted by m+1.

Therefore, in the development of any power of a binomial, the coefficients at equal distances from the two extremes are equal to each other.

REMARK. In the terms Ka"xm-n, Kam-nx", 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-n and m-n; we may therefore conclude that, the number of different combinations of m letters taken n and n, is equal to the number of combinations of m letters taken m-n and

m-n.

For example, twelve letters combined 5 and 5, give the same number of combinations as these twelve letters taken 12-5 and 12-5, or 7 and 7. Five letters combined 2 and 2, give the same number of combinations as five letters combined 5-2 and 5-2, or 3 and 3.

207. Second. If in the general formula,

m-1

(x+a)=x+max-1+m- a2xm-2+, &c.

we suppose x=1, a=1, it becomes

2

That is, the sum of the co-efficients of the different terms of the

formula for the binomial, is equal to the mth power of 2.

Thus, in the particular case

(x+a)=x+5ax2+10a2x2+10a3x2+5a*x+a,

the sum of the co-efficients 1+5+10+10+5+1 is equal to 25 or 32. In the 10th power developed, the sum of the co-efficients is equal to 210 or 1024.

208. Third. In a series of numbers decreasing by unity, of which the first term is m and the last m-p, m and p being entire numbers, the continued product of all these numbers is divisible by the con. tinued product of all the natural numbers from 1 to p+1 inclu. sively.

That is,

m(m-1) (m-2) (m-3)... (т-р)
1.2
4.... (p+1)

3

is a whole num. ber. For, from what has been said in (Art. 201), 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.

Of the Extraction of the Roots of particular numbers.

209. The third power or cube of a number, is the product arising from multiplying this number by itself twice; and the third or cube root, is a number which, being raised to the third power, will produce the proposed number.

The ten first 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 first line are the cube roots of the numbers of the second.

By inspecting these lines, we perceive that there are but nine perfect cubes among numbers expressed by one, two, or three figures; each of the other numbers has for its cube root a whole number, plus a fraction which cannot be expressed exactly by means of unity, as may be shown, by a course of reasoning entirely similar to that pursued in the latter part of (Art. 118).

210. The difference between the cubes of two consecutive numbers increases, when the numbers are increased.

Let a and a+1, be two consecutive whole numbers; we have

whence

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

(a+1)3-a3-3a2+3+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+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, its root is obtained by merely inspecting the cubes of the first nine 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 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+b2.

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

This being the case, 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 the tens must therefore be found in the part 103 which is separated from the last three 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 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, 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 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 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 into 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.

REMARK. 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.