Point every third figure, and the first period will be 15; the nearest cube to which, in the table I gave you just now, you will find to be 8, and its root 2; the 8 you must place under the 15, and the 2 in the quotient: take 8 from 15 and 7 will remain, to which bring down 6, the first figure of the next period, and you have 76 for a dividend. The figure put in the quotient is 2, the square of which is 4, which multiplied by 3 is 12, for a divisor. Now 12 in 76 will be 5 times; cube 25, and you will have 15625, which, subtract from the resolvend, and nothing will remain; which shews that the resolvend is a cube number, and 25 its root.
Pupil. You say 12 in 76 is 5 times; I should have said 6 times.
Tutor. In common division it would be so; but as the cube of 26 would be greater than the resolvend from which you are to subtract it, it can go but 5 times.
Pupil. Now, Sir, I think I have a sufficient knowledge of the rule to solve a problem.
Tutor. The earth’s period is 365 days, and its mean distance from the sun 95 millions of miles; the period of Mercury is 88 days—what is his mean distance?
Pupil. As the distance of the earth is given, I must make the square of 365 the first term, the cube of 95 the second, and the square of 88 the third term of the proportion.
Tutor. Certainly.—Take your slate, or a piece of paper, prepare your numbers, and make your proportion.
Pupil. I find the square of 365 = 133225; of 88 = 7744; and the cube of 95 = 857375.
Then 133225 : 857375 :: 7744 to a fourth term.
I now multiply the second and third terms together, and divide the product by the first, the quotient 49836 is the cube of the mean distance of Mercury from the sun in millions of miles, and the fourth term sought.