Hence, Va+3a2b+3 ab2+b3 = Va3+(3 a2+3 ab+b2)b = a + b. (1) It may be seen that the first term a of the required cube root is the cube root of the first term of the given expression arranged according to descending powers of a.

Subtracting the cube of the first term a of the root from the given expression, we obtain as a first remainder 3 ab + 3 ab2 + b3. Dividing the first term of this first remainder, arranged according to descending powers of a, by a trial divisor, 3 a2, which is obtained by taking three times the square of the first term of the root already found, we obtain the second term, b, of the root.

To obtain a complete divisor we may, as indicated in (1), add to the trial divisor, 3 a2, the term 3 ab which is three times the product obtained by multiplying the first term a of the root by the second term b, and add also the square of the second term b of the root, that is, b2.

If the complete divisor thus obtained, 3a2 + 3 ab + b2, be multiplied by b and the product be subtracted from the first remainder, 3a2b + 3 ab2 + b3, we have zero as a second remainder, and accordingly the process stops here and the required cube root is a + b. The different steps of the process are shown below:

Ex. 1. Find the cube root of 27 x6 — 54 x1y3 + 36 x2y® — 8 yo.

The process may be carried out as follows:

28. It may be shown that, by properly grouping the terms, the cube root of any polynomial cube may be obtained by repeating the steps of the process for finding the cube root of a +3 a2b+ 3 ab2 + b3. The cube of a polynomial of three terms may be arranged as follows:

(a+b+c)3 = (a+b)3 +3(a+b)2c+3(a+b)c2+c3,

}(1)

In (3) and (4) the expressions at the left of the vertical bars are the complete divisors used in the extraction of the cube root at the successive stages of the process.

29. From §§ 27 and 28 it appears that we can find the cube root of a polynomial cube of any number of terms as follows:

Rule for finding the principal cube root of a polynomial cube.

Arrange the terms of the given expression and all successive remainders according to descending or ascending powers of some letter. For the first term of the required root write the cube root of the first term of the arranged expression, and subtract its cube from the given expression to obtain the first remainder.

To obtain the next term of the root, divide the first term of the arranged first remainder by a trial divisor which is three times the square of the part of the root already found.

Construct a complete divisor by adding to the trial divisor three times the product of the part of the root previously found and the term of the root last obtained, and add also the square of the term of the root last obtained.

Subtract from the first remainder the product obtained by multiplying the complete divisor by the term of the root last obtained.

Repeat these steps, with successive remainders, until zero is obtained as a remainder.

Ex. 2. Find the cube root of 8 xo — 36 x2 + 66 xa – 63 x3 + 33 x2 - 9 x + 1. Arranged according to descending powers of x, we have:

First term of root, 8x6=2x2. | 8x6—36x5+66xa—63x3+33x2—9x+1 |2x2—3x+1

Trial divisor, 3(2x2

- 3x)2 = 12x1 — 36x3 + 27x2. +12x1-36x3

Third term of root, 12 12 1.

Complete div., 3(2xa — 3x)2 + 3(2x2 — 3x) 1 + (1)a.

(12-3633x2- 9x+1)1=

|+12xa—36x3+33x2-9x+1

EXERCISE XVIII. 5

Find the cube roots of the following expressions:

1. 27 x 54 x2 + 36 x + 8.

2. 8x+84x2y + 294 xy2+ 343 y3.

3. 512 a 1344 ab + 1176 ab2 343 b3.

4. 8a36 ab + 66 a1b2 — 63 ab3 + 33 a2b1 — 9 ab5 + bo.

5. 125a-225a+150a°+135 a5-180a++ 33a3+54a2-36a+8. 6. 343 a +441ab+777a+b2+531 ab3 +444a2b*+144 ab3+64 b. 7. 729 æo + 972 x3y + 918 x1y2 + 496 x3y3 + 204 x2y* + 48 xy* +8 yo. 8. a12-3a10-3ao + 6a2 + 8a+ 3a5-3 a1-7a3-6a2-3a-1. a as a1 3 a3 9. + + + 6 3

8

a2 За 1

+ + +

4 32 64

SQUARE ROOTS OF ARITHMETIC NUMBERS

30. Any arithmetic number may be written in the form of a polynomial whose different terms contain different powers of 10, and hence the method for extracting the square root of a polynomial may be applied to arithmetic numbers.

E. g.

6256100+ 2.10 +5

=6·102 + 2·10+5

Or,

=

410220∙ 10 + 25.

In this last form, in which the coefficients of the first and last terms are squares, the square root of 625, considered as a polynomial, may be found by the algebraic process.

The square root of 625 is thus found to be 2 ∙ 10 + 5, that is 25.

31. Consider the following relations between powers and roots :

From the relations above it may be seen that the square of an integral number having a specified number of figures may be expressed as an integral number having either twice as many figures as the given number, or as one having a number of figures one less than twice as many.

Accordingly, the number of figures in the integral part of the square root of an integral number may be found by separating the figures of the given integral number into groups of two figures each, beginning at units' place.

The number of figures in the square root will be equal to the num

ber of groups thus obtained, provided that any single figure which remains on the left is counted as a complete group.

E. g. The number of figures in the square root of 974169 is three; in the square root of 5 480281 is four.

Ex. 1. Find the square root of 3249.

Since, beginning with the units' figure 9, we can separate the figures into two groups of two figures each, 32 49, it appears that the integral part of the required square root must be a number of two figures.

Any given number expressed in the common system of arithmetic notation having two or more figures may be regarded as being composed of a certain number, t, of tens, increased by some number, u, of units. may represent any number by t⚫ 10+ u.

Hence we

Accordingly, if t· 10+u represents the number which is the required square root of the given number, 3249, we may represent the square of the square root, that is 3248,

by

(t⋅ 10+ u)2= (t. 10)2 + 2 (t⋅ 10) u + u2

=12 (100) + 2 tu· 10+ u2

Since, depending upon the value of t, t2(100) represents a number having three or four figures, it appears that for this example t must have such a value that its square t2 shall not be greater than 32.

The square next dess than 32 is 25, hence we assume that t2 represents 25, or that t = 5.

Observe that the first remainder 749 contains the two terms, 2 (t. 10) u and u2, combined into one arithmetic sum. Accordingly we cannot, as in. the algebraic process, obtain at this step the exact value of the next term of the root represented by u, by dividing 749 by 100.