"Pointing off" the number 1,728 into periods of three" (for the same reason that we point off into periods of two in square root), we proceed to extract the largest cube from the left hand period, in this case 1, the cube root of which is evidently 1 also, so that we now know we have found the 1 ten that went to make up the cube 1,728, and as the cube of the 1 ten is 1,000 we now must consider the remaining 728, which we know to be made. up 3 times 10 square times 2, plus 3 times. 10 times 2 square, plus 2 cube, as shown before. But as two is the number we are now seeking for, we must now assume that we do not know what it is, in order to be able to find it when we really do not know in actual practice.

Whether it is 2 or any other number we know it cannot be a very large number, as it must be units (as we have already found what the cube root of the tens is), and cannot therefore be over 9.

It follows then that the remaining 728 must be made up principally of 3 times 10 square, plus 3 times ten, multiplied by something which will make nearly 728,

and by using this 3 times 10 square, plus 3 times 10, for a trial divisor, we get a hint of what the "something" wanted is.

As this sum (330) will go into 728 a little over twice, we think that 2 is the number we are looking for, and to see if it really is the number we want, we must carry it through with the rest and see how

we come out.

As we know the remaining 728 to be made up of 3 times 10 square times 2, plus 3 times 10 times 2 square, plus 2 cube, we know also that if we remove the factor 2 from these quantities, that is divide them by 2, we shall obtain a quantity, which, being multiplied by 2, will give us the original quantities back again; this being self-evident.

Dividing these quantities by 2, we obtain 3 times 10 square (the 2 left out) plus 3 times 10 times 2 (one of the 2's left out) plus 2 square (one of the 2's left out again).

By making a divisor of the sum of these quantities, which amount to 364, we find it goes exactly twice into the remaining 728, and therefore know that 2 is

really the units figure that we have been looking for.

To reduce to a formula: The cube of a+b is a3+3a2b+3ab2+b3 (a=10 and b= 2 in this case), after taking away the a3 we evidently have remaining 3a2b+3ab2+ ¿3. As we do not know what b is, we take 3a2+3a, which is what remains with the b left out altogether, for a trial divisor, and by assuming the approximate quotient to be the b that is sought, we take 3a2+ 3ab+b2, which is 3a2b+3ab2+b3 divided by the b, and multiply it by the b back again, to see if the b we have assumed is really the b we have been looking after.

In the case of 1728 we find that it completes the cube, and that b is really 2 and so know that we are right.

Therefore (knowing that we are right), if there should be a remainder after taking the result of this final multiplication out of what is left of the original number, we would know that the number was not a perfect cube, and would take what is left and annex ciphers for another period, or as many more periods as required for the degree of accuracy needed, same as in square root.

Chapter XVI.

Foundation Principles.

To be thoroughly independent of rules for extracting roots (and a great many other mathematical calculations as well), we must consider the idea of factors.

When we multiply 3 by 2 we have 6 for a result, and we say 3 and 2 are factors of 6; multiplying again by 2 we get 12, and say that 6 and 2 are factors of 12, or we can say that 2 and 2 and 3 are factors of 12; or multiplying 12 by 12 we get 144 and say that 12 and 12 are factors of 144, or we can say that 2 and 2 and 2 and 2 and 3 and 3 are factors of 144, this being represented in arithmetical signs as follows: 2×2×2×2x3x3=144.

This is called an equation because something is represented as being equal to something else.

If we start and reverse the operation and divide 144 by 2 we have 72 as a result, and find that one less 2 on the left

hand side will make the equation true,

thus:

2×2×2×3×3=72.

That is, we can take away one of the factors 2 from the left hand side of this equation, and divide the right hand side by 2, and still have an equation: that is, one side will still be equal to the other side.

It follows then, that by taking one of the factors 2 from the left hand side, and dividing the right hand side by 2, we have done the same thing to both sides, because if we had not, they would no longer be equal.

That is, dividing 144 or any other number, by 2 or any other number, is simply taking the factor 2 (or whatever other number it may be), from the first number, no matter how the first number is represented.

Or in other words, taking the quantity 2×2×2×2×3×3 (which we happen to know is equal to 144) and removing one of the factors 2 from it, we have divided the whole quantity by 2.

A little more reasoning along this same line will show us that dividing a