212. If it be required to raise a" to the mth power, we shall

an expression which denotes that power of a whose index is m3. If we put m = 3, then am2 = a'.

Expressions like the above may frequently occur in algebraic operations.

EXAMPLES.

Find the value of each of the following expressions:

213. If a fraction be raised to any power by multiplication, both numerator and denominator will be raised to the same power.

Hence, to raise a fraction to any power, we have the following RULE. Raise both numerator and denominator to the required power.

=a" , and (am)n =

mn u

am Xan=am+n, where m and n are positive whole numbers. It remains to be shown that the above relations hold true when one or both of the exponents are negative. And in this investigation it is sufficient to remember that a quantity with a negative exponent is equal to the reciprocal of the same quantity with a positive exponent; (88, 2). I. To prove that am xan = am+n universally, m and n being integers.

1. Suppose one of the exponents to be negative; or let

2. Suppose both exponents are negative; or let

mm' and n == :-n'.

= amtn.

Then

am

II. To prove that

απ

= am" universally, m and n being

integers.

let

1. Suppose the exponent of the numerator to be negative; or

Then

2. Suppose the exponent of the denominator to be negative;

3. Suppose both exponents are negative; or let

mm' and n-n'.

III. To prove that (am)n = amn universally, m and n being integers.

1.-Suppose n to be negative; or let

3. Suppose both m and n to be negative; or let

m-m' and n = —n'.

Hence, in all algebraic operations, the same rules will apply to negative exponents as to positive. That is, if two powers of the same quantity be given, then the exponent of their product will be equal to the algebraic sum of the given exponents, and the exponent of their quotient will be equal to the algebraic difference of the given exponents.

EXAMPLES.

215. Find the value of each of the following expressions:

216. A polynomial may be raised to any power by actual multiplication Thus, if the quantity be multiplied by itself, the prod uct will be the second power; if the second power be multiplied by the quantity, the product will be the third power; and so on. Hence the following

RULE.-Multiply the quantity by itself in continued multiplication, till it has been taken as many times as a factor as there are units in the exponent of the required power.

NOTE. It may be well to observe that in involution we may often reach the same result by different processes. Thus, we have a6a5 × a = a4 × a2 = (a3)2 = (a2)3

4. (3a+2b+c)3.

Ans. 27a3+54a3b+27a3c+36ab2+36abc+8b3+9ac3+1263c+ 6bc3+c3.

5. (a+b)'.

Ans. a'+7ab+21a*b*+35a*b*+35a*b*+21a2b*+ 7ab'+b'. 6. (x-y)".

Ans. xo — 8x1y+28x*y* — 56x*y* + 70x*y*—56x3y* +28x3y°— 8xy' +ys.

7. (a3c ̄2+a ̄2c3)2.

Ans. a'c+2+ac.

8. (a2+1+a ̄3)3. Ans. ao+3a*+6a2+7+6a¬3+3a ̄*+a ̄°.

9. (aTM+x2)3.

Ans. am+3ax" +зax2+x3.

m

2n

POLYNOMIAL SQUARES.

217. We have seen that the square of any binomial may be written without the labor of formal multiplication, (70). Thus, if x and y represent the terms of any binomial, then

х

(x+y)2 = x2+2xy+y3.

This formula for a binomial square furnishes a simple rule for writing out the square of any polynomial, in the same direct manner. To deduce the method, let it be required to square the polynomial, a+b+c+d+e+.....

Putxa and y

=

b+c+d+e+..... Then the square will be equal to the square of the given polynomial; or

x2+2xy + y2 = (a+b+c+d+e+....)3.

And the three parts of the required square will be

x2 = a2,

2xy = 2ab+2ac+2ad+2ae+....

y2 = (b+c+d+e+........)3.

(1)

(2)

of x+y

Now y represents a polynomial; and to obtain its square, we must proceed as at first. Thus, put x' b and y'c+d+e+..... Then the square of x'+y' will be equal to the square of b+c+d+ e+..... And we have