root of 376000, to within unity, is 72; hence 47-72 =33

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12 20 20'

Find the value of 25 to within 0,001.

To do this, multiply 25 by the cube of 1000, or by 1000000000, which gives 25000000000. Now, the cube root of this number, is 2920; hence 25-2, 920 to within 0,001. (See No. 162).

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In general, in order to extract the cube root of a whole number to within a given decimal fraction, annex three times as many ciphers to the number, as there are decimal places in the required root; extract the cube root of the number thus formed to within unity, and point off from the right of this root the required number of decimals.

165. We will now explain the method for extracting the cube root of a decimal fraction. Suppose it is required to extract the cube root of 3,1415.

As the denominator 10000 of this fraction is not a perfect cube, it is necessary to make it one, by multiplying it by 100, which amounts to annexing two ciphers to the proposed decimal, and we have 3,141500. Extract the cube root of 3141500 (thatis, of the number considered independent of the comma,) to within unity; this gives 146. Then divide by 100, or ✓1000000, and we find 3.1415 1,46 to within 0.01.

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If it be required to approximate still nearer, annex three times as many ciphers more, as there are additional decimal places required in the root.

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.

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

Applying the rule of No. 162 to this example, we multiply

23 by 100, 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.

Examples.

155

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3

473, to within 20 79=4,2908, to within 0,0001 ;

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✓13=1,53, to within 0,01; √3,00415-1,4429, to within 0,00015 ✔0,00101=0,01, to within 0,01; 140,824, to within 0,001.

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

We will first consider monomials.

167. Let it be required to form the fifth power of 2a3b2. We have (No. 2)

(2a 3b3)13 =2a3b2 ×2a3b2 × 2a3b2 ×2a3b2 × 2a3b2,

from which it follows, 1st. That the coefficient 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,

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(2a3b2) 5=25.a3× 5b2×5=32a15b10. In like manner, (Ɛa2b3c)3=83.a2 ×3b3×3c3=512a©b3c3.

Therefore, in order to raise a monomial to a given power, raise the coefficient 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 coefficient. 2d. Divide the exponent of each letter by the exponent of the root.

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√64a3b3c=4a3bc2; √ 16 a® b1£ç1=2a2b3c.

1

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, its coefficient must be a perfect power, and the exponent of the letters must be divisible by the exponent or 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.

168. Hitherto we have not paid any 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, an=(a2)", every power of a quantity, of an even degree, whether positive or negative, is essentially positive.

Thus,

(+2ab3c)+16ab12c1.

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.

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Hence, (+4a2b)3 =+64ab3; (—4a3b)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 with either or

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3d. When the degree of the root is even, and the monomial uneven, the root is impossible; for there is no quantity which, raised to a power of an even degree, can give a negative result. Therefore a, b, V, are symbols of operations which it is impossible to execute, They are, like √—a, √—b, (No. 85) imaginary expressions.

169. It has already been seen in what manner a binomial x+a is raised to a power of any degree whatever; but it may happen that the terms of the binomial are affected with coefficients and exponents.

Let it be required, for example, to develope (2a2 +3ab)3.. ie 2a2=x, and 3ab=y, we have

(2a2+3ab)3(x+y)3 = x2+3x2y+3xy2+y3.

Substituting 2a2 and 3ab, for x and y, we have

(2a2 +3ab)3 =(2a2) 3 +3(2a2)3.(3ab)+3(2a2).(3ab)2 + (3ab)3, or performing the operations indicated, by the rules given in No. 167, and those for the multiplication of monomials,

(2a2+3ab)3=8a +36a5b+54a1b2 +27a3b3.

In like manner, we will find

(4a2b—3abc)=(x+y)1=x1 +4x3y+6x2y2+4xy3+y1 =(4a2b)+4(4a2b)3 (−3abc)+6(4a2 b)2 (—3abc)2 +4(4a3b) (—3abc) 3 + (—3abc)a =256a3b1-768a7b+c+864ab4c2-432a5b1c2+81a1b1c1.

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(The signs are alternately positive and negative.) In order to develope (x+y+2)3, we will take x+y=u, and we have

(u + z)3=u3 +3u2z+3uz2+z3 ;

or, replacing u by its value x+y,

(x + y + z)3 = (x+y)3+3(x+y)2.z+3(x+y)z2 +z3, and performing the operations indicated,

(x+y+z)3=x3+3x2y+3xy2+y3+3x2z+3xz2 +6xyz+3y2 z

+3yz2+z3.

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 coefficients 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+3b2) 3 = 8a — 48a5b+132a+b2-208a 3l 3

+198a2b-108ab5 +27b6.,

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

170. As to the extraction of roots of polynomials, it will be sufficient to 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, (169), 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. For it is evident, that these two terms contain the letter a, with an exponent greater than the exponent of this same letter in three times the square of the second term by the first; or in the cube of the second, three times the square of the first term by the third, &c. Hence these two parts necessarily form the first and second terms of N. Therefore, by extracting the cube root of the first term of N, we obtain the first term of R; then dividing the second term of N by 3 times the square of the first term of R, we obtain the second term of R. Knowing the two first terms of R, we can form the cube of the binomial, and subtract it from N. The remainder N' contains the product of three times the square of the first term in R, by the third, plus a series of other terms, involving a with a less exponent than it has in this product, which is consequently the first term of the remainder N'. Hence, by dividing the first term of N' by three times the square of the first term of R, we will obtain the third term of R. Cubing the trinomial found in the root, and subtracting this cube from N, we will obtain a new remainder, N", upon which we can operate in the same manner as we did upon N'; and so on.

By connecting those parts of the preceding demonstration, which are written in italics, we will form a general rule for extracting the cube root of any polynomial, and can apply it to the polynomials in No. 169,

Calculus of Radicals.

171. When it is required to extract a certain root of a mo nomial or polynomial which is not a perfect power, it can only be indicated by writing the proposed quantity after the sign √, placing over this sign the number which denotes the degree of the root to be extracted. This number is called the index of the radical.

Radical expressions may be reduced to their simplest terms