223. If the quantity which is to be involved be compound, the involution may either be represented by the proper index, or may actually be performed.

Let a+b be the quantity which is to be raised to any

Thus the square or second power of a+b is a2 + 2ab+b3, the cube or third power of a+b is a3 +3a2b+3ab2 + b3, the fourth power of a+b is a1+4a3b + 6a2b2 + 4ab3 + b1, and so on.

Similarly, the second, third, and fourth powers of a -b will be found to be respectively a2 - 2ab+b2, a3 — 3a2b+ 3ab2 — b3, and a* — 4a3b + 6a2b2 - 4ab3 +b*; that is, wherever an odd power of b occurs, the negative sign is prefixed.

We shall hereafter give a theorem, called a Binomial Theorem, which will enable us to obtain any power of a binomial expression without the labour of actual multiplication.

m

224. It is obvious that the nth power of a" is the same as the mth power of a", for each is amn; and thus we may arrive at the same result by different processes of involution. We may, for example, find the sixth power of a+b by repeated multiplication by a+b; or we may first find the cube of a+b, and then the square of this result, since the square of (a + b)3 is (a+b); or we may first find the square of a + b and then the cube of this result, since the cube of (a + b)3 is (a + b)®.

225. It may be shewn by actual multiplication that (a + b + c)2 = a2 + b2 + c3 +2ab+2bc + 2ac,

(a + b + c + d)2 = a2 + b2 + c2 + d2 +2ab+2ac + 2ad + 2bc + 2bd + 2cd. The following rule may be observed to hold good in the above and similar examples; the square of any multinomial consists of

the square of each term, together with twice the product of every pair of terms.

Another form may also be given to these results,

(a + b + c)2 = a2 + 2a (b + c) + b2 + 2bc + c2,

(a + b + c + d)2 = a2 + 2a (b + c + d) + b2 + 2b (c + d) + c2 + 2cd + ď2.

The following rule may be observed to hold good in the above and similar examples; the square of a multinomial consists of the square of each term, together with twice the product of each term by the sum of all the terms which follow it.

These rules may be strictly demonstrated by the process of mathematical induction, which will be explained hereafter.

226. The following are additional examples in which we employ the first of the two rules given in the preceding article. (a − b + c)2 = a2 + b2 + c3 — 2ab − 2bc + 2ac,

(1 − 2x + 3x2)2 = 1 + 4x2 + 9x* − 4x – 12x3 + 6x2

= 1-4x+10x2 - 12x3 +9x*,

(1 + x + x2 + x3)2 = 1 + x2 + x1 + x + 2x + 2x2 + 2x3 + 2x3 + 2x2 + 2x3

= 1 + 2x + 3x2 + 4x3 + 3x2 + 2x3 + x®.

227. The results given in Art. 55 for the cube of a + b, the cube of a-b, and the cube of a+b+c should be carefully noticed. The following may also be verified.

+3a2 (b+c+d) + 3b2 (a + c + d) + 3c2 (a + b + d) + 3d2 (a+b+c) +6bcd+6acd + 6abd + 6abc.

5. Find (1-3x+3x2 − ×3)3.

6. Shew that

(27a*18a2b-b1)
64a2b1

(9a2-b2)3 (b2 — a3)
64a2b1

= b2.

7. Shew that (ax2 + 2bxy + cy3) (aX2 + 2bXY + cY3)

= {axX+cy Y + b (xY + yX)}2 + (ac — b2) (xY — yX)3.

8. Shew that (x+ pxy + qy3) (X2 +pXY+qY2)

= (xX + pyX + qyY)3 +p(xX+pyX+qyY)(xY−yX)+q(xY−yXj2 and also

=(xX+pxY+qyY)2 +p(xX+pxY+qyY) (yX−xY) + q (xY−yX)2.

XVII. EVOLUTION.

228. Evolution, or the extraction of roots, is the method of determining a quantity, which when raised to a proposed power will produce a given quantity.

Since the nth power of a" is a"

an nth root of a"

229. mn must be a"; that is, to extract any root of a simple quantity, we divide the index of that quantity by the index of the root required.

230. If the root to be extracted be expressed by an odd number, the sign of the root will be the same as the sign of the proposed quantity, as appears by Art. 220. Thus,

231. If the root to be extracted be expressed by an even number, and the quantity proposed be positive, the root may be either positive or negative; because either a positive or negative quantity raised to an even power is positive by Art. 220. Thus,

√(a2) = ±a.

232. If the root proposed to be extracted be expressed by an even number and the sign of the proposed quantity be negative,

the root cannot be extracted; because no quantity raised to an even power can produce a negative result. Such roots are called impossible.

233. A root of a fraction may be found by taking that root of both the numerator and denominator. Thus,

234. We will now investigate the method of extracting the square root of a compound quantity.

Since the square root of a2 +2ab+b2 is a + b, we may be led to a general rule for the extraction of the square root of an algebraical expression by observing in what manner a and b may be derived from a2 + 2ab+ b2.

a2 + 2ab + b2 ( a + b

a2

2a + b) 2ab+b2

2ab+b2

Arrange the terms according to the dimensions of one letter a, then the first term is a3, and its square root is a, which is the first term of the required root. Subtract its square, that is a3, from the whole expression, and bring down the remainder 2ab+b2. Divide 2ab by 2a and the quotient is b, which is the other term of the required root. Multiply the sum of twice the first term and the second term, that is 2a + b, by the second term, that is b, and subtract the product, that is 2ab+b2, from the remainder. This finishes the operation in the present case. If there were more terms we should proceed with a+b as we did formerly with a; its square, that is a2+2ab+b2, has already been subtracted from the proposed expression, so we should divide the remainder by the double of a + b for a new term in the root, and then for a new subtrahend we should multiply this term by the sum of twice the former terms and this term. The process must be continued until the required root is found.

235. For example, required the square root of the expression 4x-12x3 + 5x2 + 6x + 1.

Here the square root of 4x* is 2x2, which is the first term of the required root. Subtract its square, that is 4x, from the whole expression, and the remainder is -1223 + 5x2 + 6x + 1. Divide - 12 by twice 2x2, that is by 4x, the quotient is - 3x, which will be the next term of the required root; then multiply 4x-3x by 3x and subtract, so that the remainder is - 4x2+6x+1. Divide by twice the portion of the root already found, that is by 4x-6x; this leads to -1; the product of 4x2 - 6x − 1 and -1 is - 4x2 + 6x + 1, and when this is subtracted there is no remainder, and thus the required root is 2x2 - 3x-1.

2x3- 6ax2 + 3α3x) 6a2x2 – 20a3ñ3 + 15a*x2 – 6a3x + ao

- 6a3x + a®