Ex. 11. To find the second power or square of any compound quantity, as a + b c + d − e + etc. If we put the quantity in the form a + (b−c+de+, etc.,) we may square it as if it were a binomial, by regarding a as one of the terms, and (b-c+de+, etc.,) as the other term of the binomial. Hence, by Ex. 9, we have (a+b-c+de+, etc.)2= a2+2abc+de+, etc.) + (b-c+de+, etc.)2= a2+2ab-c+de+, etc.) + b2 + 2b(− c +de+, etc.) +(c+de+, etc.)2= a2 + 2a(b − c + d − e +, etc.) + ―e+, b2 + 2b(− c +de+, etc.) + c2 — 2c(d — e +, etc.) + (de+, etc.)2= a2 + 2a(b − c +de+, etc.) + b2 + 2b(− c + de+ etc.) + c2 → 2c(d — e + etc.) + a2 + 2d(— e + etc.) + (-e+, etc.), and so on; as is evident since (b-cd-e + etc.) is developed as a binomial by putting b-c+d= b + (− c + d· e etc.), and in the same way, by putting -c+d c+ (d―e + etc.), we get the square of c+de+ etc., by squaring the binomial c+ (de+, etc.) by Ex. 10; and so on. e + etc. e + etc. = Hence, the square of any compound quantity is found by adding the squares of all the terms, and twice the product of each term by all the other terms connected by their proper signs; or, which is the same, we must add the squares of the terms and twice the product of every two of them without repetition; observing the rule of signs in forming the double products. For instance, to find the square of 5a2-763 + 2, we have (5a2)2 = 25a*, (— 7b2)2 = 49b*, (2c2)2 = 4c1, and of course (5a2)2 + (− 7b2)2 + (2c2)2 = 25a1 + 496* + 4c1 the sum of the squares of the terms of the given quantity. And to get the double products of every two of the terms, we have 2 × 5a2 X — 7b2 — — 70a2b3, 2 × 5a2 × 2c2 = = 20a2c2, 2 x -762 × 262 = 286; consequently, the sum of the double products is expressed by 70a2b2 + 20a3c2 — 2862c2, which, being added to the sum of the squares of the terms, gives (5a2 — 7b2 + 2c2)2 = 25a1 + 49b* + 4c1 70a2b2 + 20a2c2 286, as required. It is easy to see that the common method of squaring a number which consists of more than one digit is essentially the same as to square a compound quantity whose terms are all positive, by the rule which we have given. Thus, by squaring the number 7853, we get 7853 7853 23559 39265 62824 54971 61669609 =78532; where it will be observed that when we multiply the 3 of the multiplicand by the 3 of the multiplier, we get 9= 32, and when we multiply the 5 of the multiplicand by the 5 of the multiplier, we get 25 = 53, which is really 2500, on account of the local value of 5, which makes the 5 to be 50; in the same way, the 8 of the multiplicand multiplied by the 8 of the multiplier gives 800, and the 7 of the multiplicand multiplied by the 7 of the multiplier gives 70002; hence, the partial products in the above method of squaring 7853 contain 32, 502, 8002, and 70002. Again, the 8 of the multiplicand multiplied by 3 of the multiplier gives 800 x 3 = 2400, and the 3 of the multiplicand and multiplier being interchanged give 2 × 800 × 3; and in the same way, any figure of the multiplicand and a different figure of the multiplier will give twice the product of the figures by the numbers that express their local values. Hence, we get 78532 6166960970002 + 8002 + 50a + 32 + 2 × 7000 × 800 + 2 × 7000 × 50 + 2 × 7000.x 3 + 2 × 800 × 50 + 2 x 800 x 3 + 2 x 50 x 3; agreeably to the rule. Ex. 12. To find the cube or third power of a+a, or, which is the same, to develop (x + a) (x + a) (x + a) (x + a)3. = Since (a) = (x + a)2 (x + a) = (x2 + 2xa + a2) (x + a), we have to multiply + 2xa + a2 by (x+a). Performing the multiplication as required, we have Ex. 13. To develop (x − a) = (x − a)2(x − a) = (x2 · Ex. 14. To find the cube of any compound quantity, as a+b+c=d+e-f+, etc. From what has been shown, we have (a + b) = a3 + 3a2b +3ab2+b; and since this is an identical equation, we may for b put bc, and it becomes (a + b + c)3 = a3 + 3a2(b + c) + 3a(b + c)2 + (b + c)3 = a3 + 3a2(b + c) + 3a(b2 + 2bc + c2) + b3 + 3b3c + 3bc2 + c3 = a3 + b3 + 3a2(b + c) + 3b2(a + c) + 6abc + 3c2(a + b) + c32; and if we put in this equation C- d for c, we get (a + b + c − d)3 = a3 + b3 + 3a2(b + c d) + 3b2 (a + cd) + 6ab(cd) + 3(a + b) (c2 — 2cd+d2)+ c3cd3cd d3 = a + b3 + c3 + 3a2(b + c — d) + 36 (a + c−d) + 302 (a + b − d) + 6ab(cd) - 6cd(a + b) + 3d2(a + b + c) — d3. d+c for If, in the last equation, we put d, we shall get the expansion of (a+b+c-d+e), and so on to any extent; but it is not necessary to continue the process any further, since we easily see, from what has been done, that we can find the cube of any compound quantity by the following rule. Observing the rule of signs; the cube of any compound quantity equals the sum of the cubes of all the terms, three times the square of each term multiplied by all the others taken with their proper signs, and six times the product of every three different terms, taken with their proper signs. Thus, to get the cube of 2a-3b+4c, we have (2a)3 = 8a3, (— 3b)3 = — 2763, (4c)3 = 64c3, 3 × (2a)a ×(−3b+4c)= -36a2b+48a2c, 3(— 3b)(2a + 4c) = 54ab2 + 108b3c, 3(4c)2 (2a-3b) = 96ac2 - 144bc2, 6 × (2a) × ( — 3b) × (4c) = 144abc, and adding these results, we have (2a-3b+4c)3 = 8a3 2763 + 64c3 — 36ab + 48a2c + 54ab2 + 108b c + 96ac2 36ab+48a°c 144bc144abe, as required. = It is easy to see that the cube of a number may be found in the same way as that of a compound algebraic quantity whose terms are all positive. Thus, 3273 = 3003 + 203 + 73 + 3 x 300 x 27+ 3 x 20 x 307 + 3 x 7a × 320 + 6 × 300 × 20 × 7 = 34965783. Ex. 15.—Having the coefficients in any integral power of a binomial, to find the coefficients in the next successive integral power, by detaching the coefficients. Let a + b denote the binomial, then detaching the coefficients we have 1+ 1, and to find the coefficients in the square of a + b, we have 1+ 1 1+1 1+1 1+1 1+2+1 the coefficients in the square. To get the coefficients in the cube of a + b, we have 1+2+1 1+2+1 1+2+1 1 +3 +3 + 1 the coefficients in the cube. It is hence easily seen that if we set down the sum of the coefficients in any power in order, and then set down the same sum in the same order under the coefficients already set down, removing each term one place to the right of its equal in the first sum, then if we add the coefficients thus placed, we shall have the sum of the coefficients in the next successive integral power. Thus, 1 + 3 + 3 + 1 being the coefficients in the cube, we get the coefficients in the fourth power by writing the sum of the coefficients in the fourth power; and in like manner, to get the sum of the coefficients in the fifth power, we write 1+ 4 + 6 + 4+1 1+ 4 + 6 + 4+1 15+ 10 +10+5+1 the sum of the coefficients in the fifth power; in the same way we can find the sum of the coefficients in the sixth, seventh, etc., powers. For the purpose of raising a binomial to any integral power, which does not exceed the tenth, we shall give the following Table of corresponding coefficients. TABLE OF THE COEFFICIENTS IN THE POSITIVE INTEGRAL POW ERS OF A BINOMIAL, UP TO THE TENTH POWER. (1 + 1)1 = 1 + 1. (1 + 1)2 = 1 + 2 + 1. (1 + 1)3 = 1 + 3 + 3 + 1. The exponents of (1 + 1) in these equations indicate the powers to which the corresponding coef(1 + 1)3 = 1 + 5 + 10 + 10 + 5+ 1. ficients belong. (1 + 1) = 1 + 6 + 15 + 20 + 15 + 6 + 1. (1 + 1)* = 1 + 4 + 6 + 4 + 1. (1+1)=1+ 7 + 21 + 35 + 35 + 21 + 7 +1. (1 + 1) = 1 + 8 + 28 + 56 + 70 + 56 + 28 +8 +1. 91. (1+1)10 = 1+10 +45 + 120 +210+252 +210+120 45 + 10 +1. We shall give several examples to show the use of the Table. 1. To find the fourth power of a + b. By the Table the coefficients in the fourth power are 1 + 4 + 6 + 4 + 1; consequently, supplying the corresponding powers of the letters, we get a1 + 4a3b + 6a2b2 + 4ab3 + b=(a+b), as required. 2. To find the fourth power of 2x2 + 3y2. Here we have the same coefficients as before, and we must put 2a2 for a, and 3y2 for b; that is, we must use 23 and 3y2 in the power in the same way as we use the single letters a and b in the binomial a + b. Hence we get |