the cubes, or third powers, of x-a and x+a.

EXAMPLES FOR PRACTICE.

1. Required the cube or 3d power or 3a2.

2. Required the 4th power of 2ab.

3. Required the 3d power of - 4a2b2.

4. To find the biquadrate of

a2x

262

2x.

SIR ISAAC NEWTON'S RULE for raising a Binomial to any Power whatever*.

The index

1. To find the Terms without the Co-efficients. of the first, or leading quantity, begins with the index of the given power, and in the succeeding terms decreases continually by 1, in every term to the last; and in the 2d or

Also, powers or roots of the same quantity, are multiplied by one another, by adding their exponents; or divided, by subtracting their exponents.

a3

Thus, aз Xaз

a3+2a5. And a3a2 or

a3-3a.

42

* This rule, expressed in general terms, is as follows:

Note. The sum of the co-efficients, in every power, is equal to the

following quantity, the indices of the terms are 0, 1, 2, 3, 4, &c. increasing always by 1. That is, the first term will contain only the 1st part of the root with the same index, or of the same height as the intended power: and the last term of the series will contain only the 2d part of the given root, when raised also to the same height of the intended power: but all the other or intermediate terms will contain the products of some powers of both the members of the root, in such sort, that the powers or indices of the 1st or leading member will always decrease by 1, while those of the 2d member always increase by 1.

2. To find the Co-efficients. The first co-efficient is always 1, and the second is the same as the index of the intended power; to find the 3d co-efficient, multiply that of the 2d term by the index of the leading letter in the same term, and divide the product by 2; and so on, that is, multiply the co-efficient of the term last found by the index of the leading quantity in that term, and divide the product by the number of terms to that place, and it will give the co-efficient of the term next following; which rule will find all the co-efficients, one after another.

Note. The whole number of terms will be 1 more than the index of the given power: and when both terms of the root are +, all the terms of the power will be +; but if the second term be, all the odd terms will be +, and all the even terms —, which causes the terms to be + and — alternately. Also the sum of the two indices, in each term, is always the same number, viz. the index of the required power; and counting from the middle of the series, both ways, or towards the right and left, the indices of the two terms are the same figures at equal distances, but with mutually changed places. Moreover, the co-efficients are the same numbers at equal distances from the middle of the series, towards the right and left; so by whatever numbers they increase to the middle, by the same in the reverse order they decrease to the end.

number 2, when raised to that power. Thus 1+1=2 in the first power; 1+2+1=4=2' in the square; 1+3+3+18=2' in the cube, or third power and so on.

A trinomial or a quadrinomial may be expanded in the same manner. Thus, to raise a-b+c-d to the 6th power, put a-bx, c-d: and raise xz to the 6th power; after which substitute for the powers of x and y their corresponding values in terms of a-b, and c-d, and their powers respectively.

EXAMPLES.

1. Let a + be involved to the 5th power.

The terms without the co-efficients, by the 1st rule, will be

a", a1x, aˆx2, a2xˆ, ax1, x3,

and the co-efficients, by the 2d rule, will be

5 X 4 10 X 3 10 X 2 5 × 1

1, 5,

or, 1, 5, 10,

Therefore the 5th power altogether is

a+5a3x+10ax2 + 10a3x3 + 5ax1 + x3.

But it is best to set down both the co-efficients and the powers of the letters at once, in one line, without the intermediate lines in the above example, as in the example here below. The operation is very easily effected by performing the division first.

2. Let a x be involved to the 6th power.

α

The terms with the co-efficients will be

6ax+15a1x2-20a3x3 + 15a2x1

3. Required the 4th power of a -X.

Ans. a4a3x + 6a2x2-4ar3 + x1.

And thus any other powers may be set down at once, in the same manner, which is the best way.

4. Involve az to the ninth power; x power, and a + b -c to the fourth power.

y to the tenth

EVOLUTION.

EVOLUTION is the reverse of Involution, being the method of finding the square root, cube root, &c. of any given quan. tity, whether simple or compound.

CASE I. To find the Roots of Simple Quantities.

EXTRACT the root of the co-efficient, for the numeral part; and divide the index of the letter or letters, by the index of

the power, and it will give the root of the literal part; then annex this to the former, for the whole root sought*.

To find the square root of a Compound Quantity.

THIS is performed like as in numbers, thus:

1. Range the quantities according to the dimensions of one of the letters, and set the root of the first term in the quotient.

2. Subtract the square of the root, thus found, from the first term, and bring down the next two terms to the remainder for a dividend; and take double the root for a di. visor.

* Any even root of an affirmative quantity, may be either + or -: thus the square root of a is either+a, ora; because+ax+ a = + a2, and - -a X a = + a2 also.

But an odd root of any quantity will have the same sign as the quantity itself: thus the cube root of a3 is +a, and the cube root of isa; for+ax+ax+a=a3, and a X — a ×·

-as

Any even root of a negative quantity is impossible; for neither + a

x+a, nora X — a can produce - 02.

Any root of a product is equal to the like root of each of the factors multiplied together. For the root of a fraction, take the root of the numerator, and the root of the denominator.

3. Divide the dividend by the divisor, and annex the result both to the quotient and to the divisor.

4. Multiply the divisor, thus increased, by the term last set in the quotient, and subtract the product from the dividend.

And so on, always the same, as in common arithmetic.

EXAMPLES.

1. Extract the square root of a1· · 4a3b+6a2b2 — 4ab3+ba. a2-4a3b + 6a2ba 2ab+b2 the root.

-4ab3 + b2 (a2

2. Find the root of a + 4a3b + 10a2b3 + 12ab3 + 9b1. a'+4ab10ab2 + 12ab3 + 9b1 (a2 + 2ab+362.

a'.

2a2+2ab) 4a3b + 10a2b2

4a3b+4a2b2

2a+4ab3b2) 6a2b2 + 12ab3 + 9b1

6a2b2+12ab3 + 9b1

3. To find the square root of aa + 4a3 + 6a2 + 4a + 1.

Ans. a2+2a + 1. 2a3+2a2 a+i.

5. It is required to find the square root of a2- ab.

Ans. a2

a + 1.

b ba b3

CASE II.

To find the Roots of any Powers in general.

THIS is also done like the same roots in numbers, thus: Find the root of the first term, and set it in the quotient. -Subtract its power from that term, and bring down the second term for a dividend.--Involve the root, last found, to the next lower power, and multiply it by the index of the