Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

the product, after advancing it two places to the right, to the last term of the 2d column. Again, multiply this last result by this second figure of the root, and subtract the product from the div idend, and to the remainder annex the next period, for a NEW

DIVIDEND.

Proceed with this second figure of the root, precisely as was done with the first figure; and so continue until all the periods have been brought down.

[blocks in formation]

The greatest cube of the first period, 387, is 343, whose root is 7, which we place to the right of the number, for the first figure of the root sought. We also place it for the first term of the first column, which, multiplied into itself, gives 7×7 =49, for the first term of the 2d column, which, in turn, multiplied by 7, gives 49×7=343, which, subtracted from the first period, 387, leaves the remainder 44, to which, annexing the next period, 420, we get 44420, for the first dividend.

Again, adding 7 to the first term, 7, of 1st column, we get 7+7=14, for the second term of the 1st column, which, mul

tiplied by 7, gives 14x7=98; this, added to the first term of the 2d column, gives 147, for the second term of the 2d column, or the first trial divisor.

Again, adding 7 to the second term of the 1st column, we get 14+7=21, for the third term of the 1st column.

The trial divisor, with two ciphers annexed, becomes 14700, which is contained 3 times in the first dividend, 44420. But the trial divisor being less than the true divisor, it will sometimes give too large a quotient figure; such is the case in this present example, where 2 is the second figure of the root.

This second figure 2 of the root, annexed to the third term of the 1st column, gives 212, which, multiplied by 2, gives 424, which, being advanced two places to the right, must be added to 147, the last term of the 2d column. The sum 15124 will form the third term of the 2d column, which, multiplied by 2, gives 15124×2=30248, which, subtracted from the first dividend, leaves 14172, for the remainder, to which annexing the next period, 489, we get 14172489, for the second dividend.

Again, to the last term, 212, of the 1st column, adding 2, we get 214 for the next term, which, multiplied by 2, gives 428, which, added to 15124, gives 15552, for the second trial divisor. Again, adding 2 to 214, we get 216 for the fifth term of the 1st column.

The second trial divisor, with two ciphers annexed, becomes 1555200, which is contained 9 times in the second dividend, 14172489; therefore 9 is the third figure of the root, which, annexed to 216, givès 2169 for the last term of the 1st column, which, multiplied by 9, gives 19521, which, advanced two places to the right, and then added to 15552, gives 1574721; this multiplied by 9, gives 14172489, which, subtracted from the second dividend, leaves no remaider.

[blocks in formation]

3. What is the cube root of 10077696 ?

4. What is the cube root of 2357947691? 5. What is the cube root of 42875?

6. What is the cube root of 117649 ?

Ans. 35.

Ans. 49.

7. What is the cube root of 350356403707485209?

Ans. 704969.

8. What is the cube root of 75084686279296875?

254171347389

25419728251899

25419728251899

0

Ans. 216.

Ans. 1331.

[blocks in formation]

To extract the cube root of a decimal fraction, or of a number consisting partly of a whole number and partly of a decimal value, we have this

RULE.

I. Annex ciphers to the decimals, if necessary, so that the whole number may be divisible by 3.

II. Separate the decimals into periods of three figures each, counting from the decimal point towards the right, and proceed as in whole numbers.

NOTE-If the given number has not an exact root, there will be a remainder after all the periods have been brought down. The process may be continued by annexing ciphers for new periods.

[blocks in formation]

To extract the cube root of a vulgar fraction, or mixed number, we have this

RULE.

I. Reduce the fraction, or mixed number, to its simplest fractional form.

II. Extract the cube root of the numerator and denominator separately, if they have exact roots, but when they have not, reduce the fraction to a decimal, and then extract the root by Case II.

[blocks in formation]

When there are many decimal places required in the root, we may, after obtaining one more decimal figure than half the required number, find the rest by dividing the remainder by the last term of the SECOND COLUMN.

Before dividing, we can omit from the right of the divisor so many figures as to leave but one more than the number of additional figures required in the root; observing to omit from the right of the dividend one figure less than was omitted in the divisor. The division must then be performed according to the abridged method, as explained under Art. 42, page 56.

« ΠροηγούμενηΣυνέχεια »