and also by the fraction and by the fraction unity to the sum; place of, and a X and so on; to unite all these terms, and add and lastly, to multiply the whole by the factor xm. (2 x3- 5 a3), we must write (2x3) in the tion of the formula as an exercise for the learner.† 145. We may easily reduce the development of the power of any polynomial whatever, to that of the powers of a binomial, as may be shown with respect to the trinomial a+b+c, the third power for instance being required. First, we make b + c = m, we then obtain 3 (a + b + c)3 = (a + m)3 = a3 + 3 a2 m + 3 a m2 + m3 ; substituting for m the binomial b+c, which it represents, we have = (a+b+c)3 a3 + 3 a3 (b + c) + 3 a (b + c)2 + (b + c)3. It only remains for us to develop the powers of the binomial b+c, and to perform the multiplications, which are indicated; we have then a 3 + 3 a2 b + 3 a b2 + b3 + 3 a3 c + 6 abc + 3b2 c + c3. Of the Extraction of the Roots of Compound Quantities. 146. HAVING explained the formation of the powers of compound quantities, I now pass to the extraction of their roots, beginning with the cube root of numbers. In order to extract the cube root of numbers, we must first become acquainted with the cubes of numbers, consisting of only one figure; these are given in the second line of the following table; The formula for the development of (x + a)m answers for all values of the exponent m, and is equally applicable to cases in which the exponent is fractional or negative. This property, which is very important, is demonstrated in a note to the last part of the Cambridge course of Mathematics on the Differential and Integral Calculus. and the cube of 10 being 1000, no number consisting of three figures can contain the cube of a number consisting of more than one. The cube of a number consisting of two figures is formed in a manner analogous to that, by which we arrive at the square; for if we resolve this number into tens and units, designating the first by a, and the second by b, we have (a + b)3 = a3 + 3a2 b + 3 a b2 + b3. Hence it is evident, that the cube, or third power of a number composed of tens and units, contains four parts, namely, the cube of the tens, three times the square of the tens multiplied by the units, three times the tens multiplied by the square of the units, and the cube of the units. If it were required to find the third power of 47, by making a4 tens or 40, b = 7 units, we have 103,823 64 47 48 Now to go back from the cube 103823 to its root 47, we begin by observing that 64000, the cube of the 4 tens, contains no significant figure inferior to thousands; in seeking the cube of the tens therefore, we may neglect the hundreds, the tens, and the units of the number 103823. Pursuing, therefore, a method similar to that employed in extracting the square root, we separate, by a comma, the first three figures on the right; the greatest cube contained in 103 will be the cube of the tens. It is evident from the table, that this cube is 64, the root of which is 4; we therefore put 4 in the place assigned for the root. We then subtract 64 from 103; and by the side of the remainder, 39, bring down the last three figures. The whole remainder, 39823, contains still three parts of the cube, namely, three times the square of the tens multiplied by the units, or 3 a b, three times the tens multiplied by the square of the units, or 3 a b2, and the cube of the units, or b3. If the value of the product 3 a3 b were known, we might obtain the Alg. 20 398,23 units b, by dividing this product by 3 a2, which is a known quantity, the tens being now found; but, on the supposition that the product, 3ab, is unknown, we readily perceive, that it can have no figure inferior to hundreds, since it contains the factor a3, which represents the square of the tens; it must, therefore, be found in the part 398, which remains on the left of the number 39823, after the tens and units have been separated, and which contains, besides this product, the hundreds arising from the product, 3 a b2, of the tens by the square of the units, and from the cube b3, of the units. If we divide 398 by 48, which is triple the square of the tens, 3 a2 or 3 X 16, we obtain 8 for the quotient; but from what precedes, it appears that we ought not to adopt this figure for the units of the root sought, until we have made trial of it, by employing it in forming the three last parts of the cube, which must be contained in the remainder 39823. Making = 8, we find As this result exceeds 39823, it is evident that the number 8 is too great for the units of the root. If we make a similar trial with 7, we find that it answers to the above conditions; 47 therefore is the root sought. Instead of verifying the last figure of the root in the manner above described, we may raise the whole number expressed by the two figures, immediately to a cube; and this last method is generally preferred to the other. Taking the number 48 and proceeding thus, we find As the result is greater than the proposed number, it is evident, that the figure 8 is too large. 147. What we have laid down in the above example may be applied to all cases, where the proposed number consists of more than three figures and less than seven. Having separated the first three figures on the right, we seek the greatest cube in the part, which remains on the left, and write its root in the usual place; we subtract this cube from the number to which it relates, and to the remainder bring down the last three figures; separating now the tens and the units, we proceed to divide what remains on the left, by three times the square of the tens found; but before writing down the quotient as a part of the root, we verify it by raising to the cube the number consisting of the tens known, together with this figure under trial. If the result of this operation is too great, the figure for the units is to be diminished; we then proceed in the same manner with a less figure, and so on, until a root is found, the cube of which is equal to the proposed number, or is the greatest contained in this number, if it does not admit of an exact root. As we have often remainders, that are very considerable, I will here add to what has been said, a method, by which it may be soon discovered, whether or not the unit figure of the root be too small. or The cube of a + b, when b = 1, becomes that of a + 1, a3 +3a2 + 3 a + 1, a quantity, which exceeds a3, the cube of a, by 3a2 + 3 a + 1. Hence it follows, that whenever the remainder, after the cube root has been extracted, is less than three times the square of the root, plus three times the root, plus unity, this root is not too small. 148. In order to extract the root of 105823817, it may be observed, that whatever be the number of figures in this root, if we resolve it into units and tens, the cube of the tens cannot enter into the last three figures on the right, and must consequently be found in 105823. But the greatest cube contained in 105823 must have more than one figure for its root; this root then may be resolved into units and tens, and, as the cube of the tens has no figure inferior to thousands, it cannot enter into the three last figures 823. If, after these are separated, there remain more than three figures on the left, we may repeat the reasoning just employed, and thus, dividing the number proposed into portions of three figures each, proceeding from right to left, and observing that the last portion may contain less than three figures, we come at length to the place occupied by the cube of the units of the highest order in the root sought. Having thus taken the preparatory steps, we seek, by the rule given in the preceding article, the cube root of the two first por 64 41 8,23 The 103 823 2 0008,17 105 823 817 48 6627 tions on the left, and find for the result 47; 105,823,817 473 we subtract the cube of this number from the two first portions, and to the remainder 2000 bring down the following portion 817. number 2000817 will then contain the three last parts of the cube of a number, the tens of which are 47, and the units remain to be found. 000 000 000 These units are therefore obtained as in the example given in the preceding article, by separating the two last figures on the right of the remainder, and dividing the part on the left by 6627, triple the square of 47. Then making trial with the quotient 3, arising from this division, by raising 473 to a cube, we obtain for the result the proposed number, since this number is a perfect cube. The explanation, we have given, of the above example, may take the place of a general rule. If the number proposed had contained another portion, we should have continued the operation, as we have done for the third; and it is to be recollected always, that a cipher must be placed in the root, if the number to be divided on the left of the remainder happen not to contain the number used as a divisor; we should then bring down the following portion, and proceed with it, as with the preceding. 149. Since the cube of a fraction is found by multiplying this frac tion by its square, or which amounts to the same thing, by taking the cube of the numerator and that of the denominator; reversing this process, we arrive at the root, by extracting the root of the new numerator and that of the new denominator. The cube of §, ample, is ; taking the cube root of 125 and of 216, we find §. for ex We always proceed in this way, when the numerator and denominator are perfect cubes; but when this is not the case, we may avoid the necessity of extracting the root of the denominator, by multiplying the two terms of the proposed fraction by the square of this denominator. The denominator thence arising, will be the cube of the original denominator; and it will be only necessary then to find the root of the numerator. If we have, for example,, by multiplying the two terms of this fraction by 25, the square of the denominator, we obtain 75 5 X 5 X 5 The root of the denominator is 5; while that of 75 lies between |