Substituting these quantities for a, b, n in the foregoing general formula, it appears, that The first}.. (a") 2d 3d term 4th. ... (na"-16) (n(n−1) avabo) . . . . 2 is (x2)5 is 5× (x2) × 3y =15x3y2. 4 is 5× 2 × (x2)3 × (3y3)3 =90x°y*. 4 × × (x*)* × (3y3)3 =270x*y°. (n(n−1) (n—2) ar—sl3) is 5× 2.3 33 2 3 4 3 2 (n − 1) (n − 2) (n −3) arsb) is 5××××× (3y2)* =405x°y®. Last (bm) : is (3y2)5=243y1o. So that (x2+3y3)5=x1+15x3y2+90xoy*+270x1y® +405x3y®+243y", In the application of this formula, it may be observed, that the number of terms of which the binomial consists, is always one more than the index of the given power; after having calculated therefore as many terms as there are units in the index of the given power, we may immediately proceed to the last term. Ex. 2. Raise 3x+2y to the 6th power. Here 3x=a 2y=b n=6 and (3x+2y)=729x+2916x3y+4860x*y* Here xa and comparing (x-2y3) with (á — b)*, we have 2y=b> x2 - 14x3y2 + 84x3y1 — 280x1y® + 560x3y® — n=7) 672x2y'+448xy-128y" for the quantity re quired. 54. By means of this Theorem, we are enabled to raise a trinomial or quadrinomial quantity to any power, without the process of actual multiplication. Thus, suppose it was required to to square a+b+c; inclosing it in a parenthesis (a+b), and con- In the same manner we have, Ex. 1. (a+b+c+d)'=(a+b)+(c+d)=(a+b)2+2(a+b) (c+d)+(c+d)'=a'+2ab+b2+2ac+2ad+2bc+2bd+c2 +2cd+d2= a2+b2 + c2+ d2 + 2 (ab+ac+ad+bc+bd+cd). Ex.2. (a+b+c)3 = (a+b) + c)3 = (a+b)3 +3(a+b)°c + 3(a+b)c3+c3=a3+3a2b+3ab2+b3+3a2c+6abc+ 3 b°c + 3 ac2 +3bc'+c3= a3+b3+c3+3(a2b+ab2 + a°c + ac2 + b2c + bc3) +6abc=a3+b3 + c3 + 3 (a+b+c) (ab+ac+bc)−3abc. Ex.3. (x+y+3x)'= (x + y) + 3 z)' = (x+y)°+ 2(x+ý)× 3x+(3x) = x2+2xy+y2+6xz+6yx+9x2. XIV. On the Evolution of Algebraic Quantities. 55. Evolution," or the Rule for extracting the root of any quantity," is just the reverse of Involution; and to perform the operation, we must inquire what quantity multiplied into itself, till the number of factors amount to the number of units in the index of the given root, will generate the quantity whose root is to be extracted. 56. This rule, as applied to small numbers and simple algebraic 49=7x7; .. the square root of 49, or /49=7. 3 — b3——b x−b x−b; ;. the cube root of — b3— (√ —¿3) — — b. 36. the 4th or biquadrate root of 16a+ 816* 62=2×2×2×2×2; .. the fifth root of 32 (32)=2. 57. If the quantity under the radical sign does not admit of resolution into the number of factors indicated by that sign, or, in other words, if it be not a complete power, then its exact root cannot be extracted, and the quantity itself, with the radical Sign annexed, is called a Surd. Thus 37, Ja3, /P, $47, &c. &c. are Surd quantities. The application of the fundamental rules of arithmetic to quantities of this kind will form the subject of Chap. VIII. 58. In the involution of negative quantities, it was observed that the even powers were all +, and the odd powers - ; there is consequently no quantity which, multiplied into itself in such manner that the number of factors shall be even, can generate a negative quantity. Hence quantities of the form Fa3, √—19, √—a3, √=5, √—a*, &c. &c. have no real root, and are therefore called impossible. 59. In extracting the roots of compound quantities, we must observe in what manner the terms of the root may be derived from those of the power. For instance, (by Art. 52,) the square of a+bis a'+2ab+b', where the terms are arranged according to the powers ofa. On comparing a+b with a'+2ab+b2, we observe that the first term of the power (a) is the square of the first term of the root (a). Put a therefore for the first term of the root, square it, and subtract that square from the first term of the power. Bring down the other two terms' 2ab+b, and double the first a2+2ab+b2 (a+b a2 2a+b2ab+b2 * term of the root; set down 2 a, and having divided the first term of the remainder (2 ab) by it, it gives b, the other term of the root; and since 2ab+b2=(2a+b)b, if to 2a the term b is added, and this sum multiplied by b, the result, is 2ab+b; which being subtracted from the two terms brought down, nothing remains. 60. Again, a2+2ab+b2+2ac+2bc+c2(a+b+c a2 60. Again, the square of a+b+c (Art. 54.) is a+2ab+b2 +2ac+2bc + c'; in this case the root may be derived from the power, by continuing the process in the last Article. Thus, having found the two first terms (a+b) of the root as before, we bring down the remain 2a+b2ab+b2 2ab+b2 2a+2b+c2ac+2bc+c2 [2ac+2bc+c2 * ing three terms 2ac+2bc+c of the power, and dividing 2ue by 2 a, it gives c, the third term of the root. Next, let the last term (b) of the preceding divisor be doubled, and add c to the divisor thus increased, and it becomes 2a +2b+c; multiply this new divisor by c, and it gives 2ac+2bc+c2, which being subtracted from the three terms last brought down, leaves no remainder. In this manner the following Examples are solved. Ex. 2. x+4x+2x2+9x;2-4x+4(x3+2x2−x+2 2x2+4x3-2x+2)+4x3+8x2-4x+4 +4x+8x-4x+4 61. The process for extracting the Cube Root of a compound quantity may be explained in the following manner. Art. 52. the cube of a+b is a + 3a2b+3ab+b3, the terms being arranged accord ing to the powers of a. The first By term of the root is a, which being cubed, and this cube subtracted from the first term in the power (a3), bring down the remaining three terms 3 ab+3ab2+b3. Next square the first term (a) of the root, and having multiplied it by 3, place 3 a'in the divisor, divide 3 a'b by 3 a", and it gives b the second term of the root; to 3a add 3 ab+b, and it forms the divisor 3a2+3ab+b2, which being multiplied by b gives 3 ab+3ab+b3; subtract, and nothing remains. 62. The cube root of a compound quantity, if that root consists of three terms, is found by continuing the process in a similar manner. (a+b)3 +3(a+b)°c + 3 (a + b) c2 + c′ (a+b+c (a+b)3 3(a+b)+3(a+b)c+c2 3 (a+b)°c + 3 (a + b) c2 + c3 3 (a+b)°c + 3 (a+b) c2+c3 Thus (by Art. 54) the cube of a+b+c is (a+b)3+3(a+b)°c + 3 (a+b) c2+; supposing the first two terms of the root to have been found as in the preceding article, cube a+band subtract (a+b)3 from the first term of the power; and then bring down the next three terms 3(a+b)*c+3(a+b)c2+c2. Square the two terms already found; which square being multiplied by 3, gives 3(a+b)'; divide 3(a+b)c by 3(a+b), and we have c, the third term of the root. To 3(a+b) add 3(a+b)c+c, and it forms the divisor 3(a+b)' |