Hence, to extract the cube root of a decimal number, we have the following

RULE

Annex ciphers to the decimal part, if necessary, until it can be divided into exact periods of three figures each, observing that the number of periods must be made equal to the number of decimal places required in the root. Then, extract the root as in entire numbers, and point off as many places for decimals as there are periods in the decimal part of the number.

To extract the cube root of a vulgar fraction to within a given decimal fraction, the most simple method is to reduce the proposed fraction to a decimal fraction, continuing the operation until the number of decimal places is equal to three times the number required in the root. The question is then reduced to extracting the cube root of a decimal fraction,

219. Suppose it is required to find the sixth root of 23, to within 0,01.

Applying the rule of Art, 215 to this example, we multiply 23 by 100o, or annex twelve ciphers to 23, extract the sixth root of the number thus formed to within unity, and divide this root by 100, or point off two decimals on the right.

In this way we will find that √23=1,68, to within 0,01,

Formation of Powers and Extraction of Roots of Algebraic Quantities. Calculus of Radicals.

We will first consider monomials.

220. Let it be required to form the fifth power of 2a3b2. We

have

(2a3b2)5=2a3b32a3b3×2a3b3×2a3b3×2a3b2,

from which it follows, 1st. That the co-efficient 2 must be multiplied by itself four times, or raised to the fifth power. 2d. That each of the exponents of the letters must be added to itself four times, or multiplied by 5.

Hence,

In like manner,

(8a b3c)3=83.a3×33×3c3-512ab9c3.

'Therefore, in order to raise a monomial to a given power, raise the co-efficient to this power, and multiply the exponent of each of the letters by the exponent of the power.

Hence, reciprocally, to extract any root of a monomial, 1st. Extract the root of the co-efficient. 2d. Divide the exponent of each letter by the index of the root.

From this rule, we perceive, that in order that a monomial may be a perfect power of the degree of the root to be extracted, 1st. its co-efficient must be a perfect power; and 2d. the exponent of each letter must be divisible by the index of the root to be extracted. It will be shown hereafter, how the expression for the root of a quantity which is not a perfect power is reduced to its simplest

terms.

221. Hitherto, we have paid no attention to the sign with which the monomial may be affected; but if we observe, that whatever may be the sign of a monomial, its square is always positive, and that every power of an even degree, 2n, can be considered as the nth power of the square, that is, a2=(a2)", it will follow that,

every power of a quantity, of an even degree, whether positive or negative, is essentially positive.

Again, as a power of an uneven degree, 2n+1, is the product of a power of an even degree, 2n, by the first power, it follows that, every power of an uneven degree of a monomial, is affected with the same sign as the monomial.

Hence, (+4a3b)3=+64ab3; (-4a2b)3-64ab3.

From this it is evident, 1st. That when the degree of the root of a monomial is uneven, the root will be affected with the same sign as the quantity.

2d. When the degree of the root is even, and the monomial a positive quantity, the root is affected either with + or -.

Thus,

V81a4b123ab3;

√64a13±2a3.

3d. When the degree of the root is even, and the monomial negative, the root is impossible; for, there is no quantity which, raised to a power of an even degree, can give a negative result. Therefore, V-a, √ b, c, are symbols of operations which it is impossible to execute. They are, like √—a, √―b, imaginary expressions (Art. 126).

8

222. In order to develop (a+y+z)3, we will place y+z-u, and we have

(a+u)3=a3+3a2u+3au2+u3,

or by replacing u by its value, y+z

(a+y+z)3=a3+3a2(y+z)+3a(y+2)2+(y+z)3,

or performing the operations indicated

(a+y+z)3=a3+3a2y+3a3z+3ay2+6ayz+3az2+y3+3y2z+

3yz2+z3.

When the polynomial is composed of more than three terms, as

... •

a+y+z+x p, let, as before, u= the sum of all the terms after the first. Then, a+u will be equal to the given polynomial, and

(a+u)3=a3+3a3u+3au2+u3.

From which we see, that by cubing a polynomial, we obtain the cube of the first term, plus three times the square of the first term multiplied by each of the remaining terms, plus other terms.

It often happens that u contains a root of the leading letter, as in the polynomial a2+ax+b, where u=ax+b. But since we suppose the polynomial arranged with reference to a, it follows that a will have a less exponent in u than in the first term.

In this case also, the co-efficient of u, multiplied by the first term of u, will be irreducible with the remaining terms of the development, because that product will involve a to a higher power than the other terms: and when a does not enter u, the product of that co-efficient by all the terms of u, will be irreducible with all the other terms of the development.

223. As to the extraction of roots of polynomials, we will first explain the method for the cube root; it will afterwards be easy to generalize.

Let N be the polynomial, and R its cube root. Conceive the two polynomials to be arranged with reference to some letter, a, for example. It results from the law of composition of the cube of a polynomial (Art. 222), that the cube of R contains two parts, which cannot be reduced with the others; these are, the cube of the first term, and three times the square of the first term by the second.

Hence, the cube root of that term of N which contains a, affected with the highest exponent, will be the first term of R: and the second term of R will be found by dividing the second term of N by three times the square of the first term of R.

If then, we form the cube of the two terms of the root already found, and subtract it from N, and divide the first term of the remainder by 3 times the square of the first term of R, the quotient will be the third term of the root. Therefore, having arranged

the terms of N, we have the following

RULE.

1. Extract the cube root of the first term.

II. Divide the second term of N by three times the square of the first term of R: the quotient will be the second term of R.

III. Having found the two first terms of R, form the cube of the binomial and subtract it from N; after which, divide the first term of the remainder by three times the square of the first term of R: the quotient will be the third term of R.

IV. Cube the three terms of the root found, and subtract the cube from N: then divide the first term of the remainder by the divisor already used: the quotient will be the fourth term of the root, and the remaining terms, if there are any, may be found in a similar manner.

EXAMPLES.

1. Extract the cube root of 2-6x+15x1—20x3+15x2-6x+1.

x-6x+15x-20x3+15x2-6x+1 | x2-2x+1

(x2-2x)3= x2-6x3+12xa— 8x3

3x1

3x-12x2+, &c.

(x2-2x+1)=x-6x+15a1-20x3+15x2-6x+1.

In this example, we first extract the cube root of ao, which gives a2, for the first term of the root. Squaring 2, and multiplying by 3, we obtain the divisor 3x: this is contained in the second term -6x5, -2x times. Then cubing the root, and subtracting, we find that the first term of the remainder 3x1, contains the divisor once. Cubing the whole root, we find the cube equal to the given polynomial.

REMARK. The rule for the extraction of the cube root is easily extended to a root with a higher index. For,

Let a+b+c+ f, be any polynomial.

Let s the sum of all the terms after the first.

Then as the given polynomial: and

(a+s)"a"+na"-'s+ other terms