EVOLUTION.] ALGEBRA 527 a-x being the root, or first power ; then a- - 2ax + x 2 is the square, or second power ; a - 3a 2 A- + Sax* - x 3 , the cube, or third power ; u 4 - 4a 3 # + 6a 2 x 2 - -laa? + x*, the fourth power. Hence it appears that the powers of a + x differ from the powers of a x only in this respect, that in the former the signs of the terms are all positive, but in the latter they are positive and negative alternately. Besides the method of finding the powers of a compound quantity by multiplication, which we have just now ex plained, there is another more general, as well as more expeditious, by which a quantity may be raised to any power whatever without the trouble of finding any of the inferior powers, namely, by means of what is commonly called the binomial theorem, to be proved hereafter. This theorem may be expressed as follows : Let a + x be a binomial quantity, which is to be raised to any power denoted by the number n, then (a + x) n =
- +
+ 2 (-!) n _o o tt(tt-l)(n-2) a * 1 . 2 3 This series will always terminate when n is any whole positive number, by reason of some one of the factors -/i-l, n-2, ifec., becoming = ; but if n be either a nega tive or fractional number, the series will consist of an infinite number of terms. As, however, we mean to treat in this section only of the powers of quantities when their exponents are whole positive numbers, we shall make no further remarks upon any other. The nth power of a x will not differ from the same power of a + x, except in the signs of the terms which compose it, for it will stand thus : (a x) n = _ n(n-l)(M-2) n _ 3 3 ~ 1 . 2 . 3 * X , n a a (n 1) A 1 &c., where the signs are + n(tt-l)(- 2X^-3) _. + l".2 . 3 4 a and alternately. Let it be required, for instance, to raise a + x to the fifth power. Here n, the exponent of the power, being 5, the first term a" of the general theorem will be equal to a 5 , the 5x4 second na n ^x = 5a*x, the third ~ ^ a* 1.2 3 the fourth 1x2 5x4x3 1x2x3 " "" 5x4x3x2 1x2x3x4 ,3.2 _ ^ o , ,1 ,,, - lOa-itf 3 , the fifth,- 1.2.3.4 = 5ax*, and the sixth and last <-l)(-2)(n-3)(n-4) , 5x4x3x2x1 1.2.3.4.5 "Ix2x3x4x5 the remaining terms of the general theorem all vanish, by reason of the factor n 5 = by which each of them is multiplied, so that we get (a + x) 5 = a^ + 5a*x + 10a 3 ;c 2 + If the quantity to be involved consists of more than tvro terms, as, if p + q r were to be raised to the second power, put p = a and q-r = b, then (p + q - r) 2 = (a + b) 2
a 2 + 2ab + b 2 =p 2 + 2p(q - r) + (q- rf, but 2p(q -r)
2pq - 2pr, and by the general theorem (q - r) 2 = q 2 - 2qr + r", therefore we get (p + q- r) 2 =p 2 + 2pq- 2pr + q 2 - 2qr + r 2 ; and by a similar method of proceeding a quantity consisting of four or more terms may be raised to any power. Additional Examples. Ex. 1. From the value of (a + x)* found in example 4, to find that of (a + b + c) 4 . From example 4 we write at once, by symmetry, (a + b + cY = a 4 + 4a 3 6 + Ga 2 b 2 + R + c 4 + 4 3 a + 66 2 c 2 + 46 3 c + 4c 3 a + 4c 3 6 where II is the series of remaining terms denoting the threo following forms, a 2 bc, b-ac, c 2 ab. Now when a, b, c are each unity, there are 81 terms (viz. 3 4 ). But the number of terms already written down (4a 3 6 being considered as 4 terms, &c.) is 45. The quantity E must consequently make up the other 36 terms, . . it can be nothing else than I2a 2 bc+l2b 2 ac+12c 2 ab. Ex. 2. (j) + q + r) 2 Cor. If p + q + r = Q; 2 + < + r 2 + 2(pq + qr + rp) = 0. Case 1. a-b + b-c + c-a = 0, gives (a - b) 2 + (b- c) 2 + (c - a} 2 + 2{(a-6)(6 - c) + (b - c)(c - a) + (c-a)(a-b)}=Q. Case 2. a(b c) + b(c a) + c(a 6) = 0, gives a-(b - c) 2 + bc - a) 2 + c 2 (a - 6) 2 + 2 {ab(b - c)(c - a) + bc(c - a)(a -b) + ca(a - 6)(6 - c)} = . Ex. 3. Prove that (x 2 - yzf + (f - xz) 3 + (z 2 - xy) 3 - 3(.r 2 - yz)(y 2 xz)(z 2 xij) is a complete square. The expression will assume symmetry if (x 2 yz}(y z xz) (z- xy], instead of being multiplied by 3, be repeated three times, each being connected with one of the cubes in turn; this gives x 2 - y.?){0 2 - yz) 2 - (/- - xz)(z 2 - xy)} + y 2 - xz) { (y 2 - xz} 2 - (x 2 - yz)(z 2 - xy)} + z 2 - xy) { (z 2 - xy) 2 - (x 2 - yz)(y 2 - xz)} = (x 2 - yz)x{x 3 + y 3 + z s - 3xyz} + &c., &c. = (X* + y 3 + z 3 - 3xyz)(x 3 + f + <?- 3xyz) . Ex. 4. Prove that (a 2 + b 2 + c 2 ) 3 + 2(ab + bc + ca) 3 - 3(a 2 + b 2 + c 2 )(ab + bc + ca) 2 = (a 3 + i 3 + c 3 - 3a6c) 2 . Combine each of the cubes with each of the products in succession, and reduce, as in the last example. Ex. 5. To find the condition that px 2 + 2qxy + ry 2 may be incapable of changing its sign through any change of sign or value of x and y. It is evident that p and r must have the same sign. Suppose it positive. By multiply ing by p, the quantity may be thrown into the form (px + qy) 2 + (pr - q 2 )y 2 , which is the sum of two positive quantities provided pr>(. The condition required is, therefore, pr>q 2 ; or as a particular case pr = (f-. Ex. 6. To find the condition that ax 2 + by 2 + cz* + 2P^2 + 2Q2X + 2Pury may be incapable of changing its sign through any change of sign or value of x, y, z. We will suppose a, b, c to be all positive, in which case the whole result is also positive. If we multiply the whole by a, we may write it under the form of a square and a supplement, viz., (ax + Qz + Ry) 2 + (ac - Q 2 ) 5 2 + (ab - B% 2 + 2(aP - QR)yz. Now as the first term of this expression is a square, it is essentially positive. Hence the required condition can be satisfied only by rendering the remainder positive. It follows that ac>Q 2 , ab>~R 2 , and (Example 5) (ac - Q 2 )(ab - R 2 ) > (aP - QH) 2 ,
i.e., abc + 2PQR >aP 2 + Z/Q 2 + cR 2 .
