Cube root in Sadratnamālā

Here we will be knowing cube root of a number as mentioned in Sadratnamālā.

Verse 18

घनमूलस्य वर्गेण त्रिघ्नेनाघनतोऽन्त्यतः ।

लब्धस्य वर्गस्त्रयादिघ्नः शोध्याश्चाद्याद् घनाद् घनः ॥ १८ ॥

Translation

(Having deducted the greatest possible cube from the last place and having kept the cube root of the number subtracted in the line of cube root), divide the second non-cube place by thrice the square of the cube root (and place the quotient on the right of the cube root kept earlier) and subtract the square of the quotient multiplied by thrice the cube root from the first non- cube place. Then subtract the cube (of the quotient) from the cube place. Repeat this until the digits are exhausted.

The places counted from right to left are called cube place, first non-cube place, second non-cube place, again cube place, first non-cube place, second non-cube place and so on.

Example: Cube root of 12812904

Right to left the cube places are marked by c and non-cube places are marked by n and n' respectively.

	n	c	n'	n	c	n'	n	c	Step details	Result
	1	2	8	1	2	9	0	4
-23 (2 - Prathamaphala)		8							Subtract the maximum possible cube (8 = 23) from the last cube place (12) . Here 2 is Prathamaphala.	2
÷ 3 X 22 = 12	12)	4	8	(3 (3 - Dvitīyaphala)					Place the digit of the next non-cube place (8) on the right of the remainder (4). Now the number is 48 and divide this by thrice the square of the first result (2) = 3 X 22 = 12.	2 3
		3	6						Subtract the above number from the maximum possible number 12 X 3 = 36. Here the quotient is 3. 3 is Dvitīyaphala. In order that the product of the thrice the first result and square of the quotient can be subtracted from the next non-cube place, we have kept the quotient above as 3 not 4.
		1	2	1					Place the digit of the next non-cube place (1) on the right of the remainder (12), Now the number is 121.
-3 X 2 X 32 = -54			5	4					Deduct thrice the first result multiplied by square of the quotient (3) = 3 X 2 X 32 = 54.
			6	7	2				Place the digit of the next cube place (2) on the right of the remainder (67). Now the number is 672.
-33				2	7				Subtract the cube of quotient (3).
÷ 3 X 232 = 1587	1587)		6	4	5	9	(4 (4 - Tṛtīyaphala)		Place the digit of the next non-cube place (9) on the right of the remainder (645), Now the number is 6459 and divide this by thrice the square of the second result (23) = 3 X 232 = 1587.	2 3 4
			6	3	4	8			Subtract the above number from the maximum possible number 1587 X 4 = 6348 Here the quotient is 4. 4 is Tṛtīyaphala.
				1	1	1	0		Place the digit of the next non-cube place (0) on the right of the remainder (111), Now the number is 1110.
-3 X 23 X 42 = -1104				1	1	0	4		Deduct thrice the second result multiplied by square of the quotient (4) = 3 X 23 X 42 = 1104.
							6	4	Place the digit of the next cube place (4) on the right of the remainder (6). Now the number is 64.
-43 = -64							6	4	Subtract the cube of quotient (4).
								0

As the remainder is zero , the cube root is exact.

Cube root of 12812904 = 234

Example: Cube root of 2628072

Right to left the cube places are marked by c and non-cube places are marked by n and n' respectively.

	c	n'	n	c	n'	n	c	Step details	Result
	2	6	2	8	0	7	2
-13 (1 - Prathamaphala)	1							Subtract the maximum possible cube (1 = 13) from the last cube place (2) . Here 1 is Prathamaphala.	1
÷ 3 X 12 = 3	1	6	(3 (3 - Dvitīyaphala)					Place the digit of the next non-cube place (6) on the right of the remainder (1). Now the number is 16 and divide this by thrice the square of the first result (1) = 3 X 12 = 3	1 3
		9						Subtract the above number from the maximum possible number 3 X 3 = 9. Here the quotient is 3. 3 is Dvitīyaphala. In order that the product of the thrice the first result and square of the quotient can be subtracted from the next non-cube place, we have kept the quotient above as 3 not 4 or 5.
		7	2					Place the digit of the next non-cube place (2) on the right of the remainder (7), Now the number is 72.
-3 X 1 X 32 = -27		2	7					Deduct thrice the first result multiplied by square of the quotient (3) = 3 X 1 X 32 = 27.
		4	5	8				Place the digit of the next cube place (8) on the right of the remainder (45). Now the number is 458.
-33			2	7				Subtract the cube of quotient (3).
÷ 3 X 132 = 507		4	3	1	0	(8 (8 - Tṛtīyaphala)		Place the digit of the next non-cube place (0) on the right of the remainder (431), Now the number is 4310 and divide this by thrice the square of the second result (13) = 3 X 132 = 507.	1 3 8
		4	0	5	6			Subtract the above number from the maximum possible number 507 X 8 = 4056 Here the quotient is 8. 8 is Tṛtīyaphala.
			2	5	4	7		Place the digit of the next non-cube place (7) on the right of the remainder (254), Now the number is 2547.
-3 X 13 X 82 = -2496			2	4	9	6		Deduct thrice the second result multiplied by square of the quotient (8) = 3 X 13 X 82 = 2496.
					5	1	2	Place the digit of the next cube place (2) on the right of the remainder (51). Now the number is 512.
-83 = -512					5	1	2	Subtract the cube of quotient (8).
							0

As the remainder is zero , the cube root is exact.

Cube root of 2628072 = 138

See Also

सद्रत्नमाला में 'घनमूल'