1. Find, by trial, a cube near to the given number, and call it the supposed cube.

2. Then as twice the supposed cube, added to the given number, is to twice the given number, added to the supposed cube, so is the root of the supposed cube, to the true root, or an approximation to it.

3. By taking the cube of the root, thus found, for the supposed cube, and repeating the operation, the root will be had to a greater degree of exactness.

EXAMPLE.

It is required to find the cube root of 54854153? Let 64000000=supposed cube, whose root is 400: Then, 64000000

54854153

2

2

182854153)69483322400(379=root nearly.

Again, let 54439939 = supposed cube, whose root is 379.

Then, 54439939

54854153

163734031)62212184855(379 958793-+-root corrected.

SECOND METHOD BY APPROXIMATION.

RULE.

1. Divide the resolvend by three times the assumed root, and reserve the quotient.

2. Subtract one twelfth part of the square of the assumed root from the quotient.

3. Extract the square root of the remainder.

4. To this root add one half of the assumed root, and the sum will be the true root, or an approximation to it; take this approximation as the assumed root, and, by repeating the process, a root farther approximated will be found, which operation may be farther repeated, as often as necessary, and the root discovered to any assigned exactness.

Note. In order to find the value of the first assumed root, in this or any other power, divide the resolvend into periods by beginning at the place of units, and including in each period, so many figures as there are units in the exponent of the root; viz. 3 figures in the cube root; 4 for the biquadrate, and so on; then, by a table of powers, or otherwise, find a figure, which (being involved to the power whose exponent is the same with that of the required root) is the nearest to the value of the first period of the resolvend at the left hand, and to that figure annex so many ciphers as there are periods remaining in the integral part of the resolvend; this figure, with the ciphers annexed, will be the assumed root, and equal to r in the theorem; and it is of no importance whether the figure thus chosen be; when involved, greater or less than the left hand period, as the theorem is the same in both cases.

1st. What is the cube root of 436036824287 ?

7000 assumed root.

3

211000)436036824287(20763658-2994

Subtract 7000×7000÷12=4083333-3333

✓16680324-9661-4084·15

Add the assumed root=3500

And it gives the approximated root= 7584.15

For the second operation, use the approximated root as the assumed one, and proceed as above.

THIRD METHOD BY APPROXIMATION.

1. Assume the root in the usual way, then multiply the square of the assumed root, by 3, and divide the resolvend by this product; to this quotient add of the assumed root, and the sum will be the true root, or an approximation to it.

2. For each succeeding operation let the last approximated root be the assumed root, and proceeding in this manner, the root may be extracted to any assigned exactness.

1st. What is the cube root of 7?

Let the assumed root be 2.

Then.2×2×3=12 the divisor.

12)7 0(583 to this add of 2-1333, &c. that is, 583+1.333= 1916 approximated root.

Now assume 1916 for the root. Then, by the second process,

the root is

7

3×1916

2+3x1916-19126, &c.

2d. What is the cube root of 9? Let 2 be the assumed root as before. Then, 12+3x2=208 the approximated root.

Now as

9

sume 2:08. Then,

2+3x208-208008, &c.

3×2.08

3d. What is the cube root of 282? Let 6 be the assumed root.Then, 6×6×3=108)282(2 611, &c. and 2611,+ of 6-6611 approximated root. Now assume 6·611, and it will be 6611×6611 X3-131-116)282(2-1507, &c. and 2 1507+ of 6611-6558 a farther approximated root.

4th. What is the cube root of 1728?-Here the assumed root is 10. Then, 10x10×3=300)1728(5·76, and 5·76+ of 10-12 426. -Now assume 12:426, then 12:426 ×12 426×3=463-216428)1728 (3-732, and 3-732+ of 12-426-12-014 a farther approximated root, and so on.

APPLICATION AND USE OF THE CUBE ROOT. 1. To find two mean proportionals between any two given numbers.

RULE.

1. Divide the greater by the less, and extract the cube root of the quotient.

2. Multiply the root, so found, by the least of the given numbers, and the product will be the least.

3. Multiply this product by the same root, and it will give the greatest.

EXAMPLES.

1st. What are the two mean proportionals between 6 and 750 ?

3

750+6=125, and ✔ 124=5. Then, 5×6=30-least, and 30X 5=150 greatest. Answer 30 and 150.

Proof. As 6: 30: 150 750.

2d. What are the two mean proportionals between 56 and 12096? Answer 336 and 2016. Note. The solid contents of similar figures are in proportion to each other, as the cubes of their similar sides or diameters. 3d. If a bullet 6 inches diameter weigh 32; What will a bullet of the same metal weigh, whose diameter is 3 inches?

6×6×6=216. 3x3x3=27. As 216 32: 27: 4, Ans. 4th. If a globe of silver of 3 inches diameter, be worth £45, What is the value of another globe, of a foot diameter ?

Ans. £2880. The side of a cube being given, to find the side of that cube which shall be double, triple, &c. in quantity to the given cube.

RULE.

Cube your given side, and multiply it by the given proportion between the given and required cube, and the cube root of the product will be the side sought.

5th. If a cube of silver, whose side is 4 inches, be worth £50, I demand the side of a cube of the like silver, whose value shall be 4 times as much?

3

4X4X4-64, and 64x4-256. v256-6-349+inches, Ans. 6th. There is a cubical vessel, whose side is 2 feet; I demand the side of a vessel, which shall contain three times as much? Ans. 2ft. 10 inches. 7th. The diameter of a bushel measure being 18 inches, and the height 8 inches, I demand the side of a cubic box, which shall contain that quantity. Ans. 12.907+inches.

* As two mean proportionals are required to two given numbers, there will be four terms in the proportion, in which the first is to the second, as the second to the third, and the third to the fourth. The numbers therefore belong to a geometrical progression of four terms. The first part of the rule is explained in Prob. VIII. of Geometrical Progression, and the second and third parts of the rule are evident from the proof of Prob. I. of Geometrical Progression.

The solid, called a cube, has its length and breadth and height all equal. As the number of solid feet, inches, &c. in a cube are found by multiplying the height and length and breadth together, that is, by multiplying one side into itself twice, the third power of a number is called the cube of that number.

+ Multiply the square of the diameter by 7854, and the product by the height; the cube root of the last product is the answer. See Mensuration of Superficies and Solids, Art. 30.

8. Suppose a ship of 500 tons has 89 feet keel, 36 feet beam, and is 16 feet deep in the hold: What are the dimensions of a ship of 200 tons, of the same mould and shape?

89x89x89=704969-cubed keel.

As 500 200 :: 704969: 281987.6 cube of the required keel.

√281987-6=65 57 feet the required keel.

As 89: 65 57 :: 36 : 26·522=26 feet beam, nearly.

As 89: 65 57: 16: 117 feet, depth of the hold, nearly.

9. From the proof of any cable to find the strength of any other. RULE. The strength of cables, and consequently the weights of their anchors, are as the cubes of their peripheries.

If a cable, 12 inches about, require an anchor of 18cwt: Of what weight must an anchor be, for a 15 inch cable?

Cwt.

Cwt.

As 12x12×12 : 18 :: 15×15×15: 35 15625 Ans. 10. If a 15 inch cable require an anchor 3515625cwt. · What must the circumference of a cable be, for an anchor of 18cwt?

12 inches, Answer.

EXTRACTION OF THE BIQUADRATE ROOT.

RULE.

Extract the square root of the resolvend, and then the square root of that root, and you will have the biquadrate root.

What is the biquadrate root of 20736?

1. Divide the resolvend by six times the square of the assumed root, and from the quotient subtract part of the square of the assumed root.

2. Extract the square root of the remainder.

3. Add of the assumed root to the square root, and the sum will be the true root, or an approximation to it.

4. For every succeeding operation, either in this or the following method, proceed in the same manner, as in the first, each time using the last approximated root for the assumed root.

The biquadrate root of 20736 is required.

Here 10 is the assumed root.