1. Cube 45 by means of the cubical blocks.

increase its dimen-
sions by 5 units we
must add, 1st, the
three rectangular
slabs, B, C, D, Fig.
2; 2d, the three
corner pieces, E, F,
G, Fig. 3; 3d, the
little cube H, Fig.
4. The three slabs
B, C, D, are 40 units
long and wide and
5 units thick; hence
their contents are
402X5X3=24000;
the contents of the
corner pieces, E, F,
G, Fig. 3, whose

length is 40 and breadth and thickness 5, equal 40 X 52 3=3000, and the contents of the little cube, H, Fig. 4, equal 53-125; hence the contents of the cube represented by Fig. 4 are 64000 + 24000 + 3000+ 125 91125. Therefore, the cube of 45, etc.

NOTE. When there are three figures in the number, complete the second cube as above, and then make additions and complete the third in the same manner. If there are still some figures and no more blocks to make additions, let the first cube represent the cube already found, and then proceed as at first.

Cube the following numbers geometrically :-

GEOMETRICAL METHOD OF SQUARE ROOT.

439. The Geometrical Method is so called because it makes use of a geometrical figure to explain the process of extracting the root.

1. Extract the square root of 1225.

SOLUTION. This problem by the geometrical method requires us to find the side of a square whose area is 1225 square units. Since the square of a number consists of

OPERATION.

1225(30 302- 900 5

325 35

30 X 2=60
(60+5)X5= 325

twice as many places as OPERATION AS IN PRACTICE. the number itself, or twice

1225(35

as many less one, the

3

9

square root will consist of two places, and hence will consist of tens and units.

60

5

Fig. 3.

B

D

A

The greatest number of tens whose square is contained in the given number is 3 tens. Let A, Fig. 1, represent a square whose sides are 3 tens or 30 units, its area will be the square of 30, or 900. Subtracting 900 from 1225 we find our square is not large enough by 325 square units, we must therefore increase it by 325 units. To do this we add the two rectangles B and C, Fig. 2, each of which is 30 units long, and since they nearly complete the square their area must be nearly 325 units, hence if we divide 325 by their length we will find their width. The length of each is 30, hence the length of both is 30 X 2 60, and dividing 325 by 60 we find their width to be 5 units. We now complete the square by the addition of the little corner square, D, Fig. 3, whose sides are 5 units, and then the entire length of the additions is 60+ 5 = 65 units, and multiplying this by the width we find the whole area of the additions to be 65 X 5 = 325 square units. Subtracting and nothing remains, therefore the side of the square whose area is 1225 units is 35 units, hence the square root of 1225 is 35.

The rule is the same as that already given. Let the pupil apply the method to some of the problems under the Analytic

Method.

GEOMETRICAL METHOD OF CUBE ROOT.

440. The Geometrical Method of cube root is so art because it makes use of a cube to explain the process.

1. Extract the cube root of 74088.

SOL.-We find the number of figures in the root as before, and then proceed as follows: The greatest number of tens whose cube is contained in the given number is 4 tens. Let A, Fig. 1,

represent a cube whose

Fig. 1.

Fig. 2.

sides are 40, its contents will be 403

=64000. Subtracting from 74088 we find a remainder of 10088 cubit units, hence the cube A is not large enough to contain 74088 by 10088 cubit units, we will therefore increase it by 10088 cubit units.

To do this we add the three rectangular slabs B, C, D, Fig. 2, each of which is 40 units in length and breadth, and since they nearly complete the cube their contents must be nearly 10088, hence if we

divide 10088 by the sum of the areas of one of their faces as a base, wo can ascertain their thickness.

The area of a face of one slab is 4021600, and of the three, 3 X 1600 4800, and dividing 10088 by 4800 we have a quotient of 2, hence the thickness of the additions is 2 units. We now add the three corner pieces, E, F, and G, each of which is 40 units long, 2 wide, and 2 thick, hence the surface of a face of each is 40 X 280 square units, and of the three it is 80 X 3=240 square units.

We now add the little corner cube H, Fig. 4, whose sides are and the surface of a face is 224. We now take the sum of the surunits, faces of the additions, and multiply this by the common thickness, which is 2, and we have their solid contents equal to (4800 +240 +4)

X2=10088. Subtracting, nothing remains, hence the cube which contains 74088 cubic units is 40+2 or 42 units on a side.

NOTE. When there are more than three figures we increase the size of the new cube, Fig. 4, as we did the first, or let the first cube, Fig. 1, represent the new cube, and proceed as before.

The rule is the same as that already given. Let the pupil apply this method of explanation to some of the problems given under the Analytic Method. The operation is abbreviated in practice, as is shown in the Analytic Method.

SOME PRINCIPLES OF INVOLUTION AND EVOLUTION.

441. The pupil will now give attention to the following principles of involution and evolution; with quite young pupils they may be deferred until review.

PRIN. I. The product of any two powers of a number equals a power of the number denoted by the sum of the exponents. DEM.—If we multiply the square of a number by the cube of the number, we will have the number used five times as a factor, or the 5th power of the number, and the same may be shown in any other case.

EXAMPLES FOR PRACTICE.

1. Find the fifth power of 4.

SOLUTION.-Cube 4, which is 64, and multiply this by the square of 4, which is 16, and we will have the fifth power of 4.

4. 74X75.

5. 123×125.

6. (1)2×(1)3.

Ans. 85.

Ans 79. 8 (1)3×(1)'.

Ans. 4o.

Ans (4).

Ans 128. 9. (2.5)*X(2.5). Ans (25)1o.

Ans. (1)5 10. (3.3)2X(3.3) Ans. (3.3).

PRIN. II.—A power of a number raised to any power equals a power of the number denoted by the product of the exponents

DEM.-If we square the cube of a number we will use the number as a factor two times 3 times, or 6 times; thus, (43)2 = 43 × 43—46, and the

same can be shown in any other case.

CTICE.

EXAMPLES FOR PRACTICE.

1. Find the sixth power of 5.

SOLUTION -Cube 5 and we have 125, and multiply this by itself and we have the 6th power of 5, which we find is 16625; hence the sixth power of 5 is 16625

PRİN. III.—A root of a number equals a root of a root of the number, in which the product of the indices of the two latter roots equals the index of the former.

DEM. Since the square of the cube of a number equals the sixth power, the sixth root of a number equals the square root of the cube root of the number, and the same is true in any other case.

EXAMPLES FOR PRACTICE.

1. Extract the sixth root of 4096.

SOLUTION. To find the sixth root of 4096 we first extract the square root, which we find to be 64, and then find the cube root of 64, which is 4. Hence the sixth root of 4096 is 4.