Cube root in Pāṭīgaṇitam

Introduction

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

Verse

घनपदमघनपदे द्वे घन (पद) तोऽपास्य घनमदो मूलम् ।

संयोज्य तृतीयपदस्याघस्तदनष्टवर्गेण ॥ २९ ॥

एकस्थानोनतया शेषं त्रिगुणेन (सं)भजेत्तस्मात् ।

लब्धं निवेश्य पङ्क्त्यां तद्वर्गं त्रिगुणमन्त्यहतम् ॥ ३० ॥

जह्यादुपरिमराशेः प्राग्वद् घनमादिमस्य (च) स्वपदात् ।

भूयस्तृतीयपदस्याघ इत्यादिकविधिर्मूलम् ॥ ३१ ॥

Translation

(Divide the digits beginning with the units' place into periods of) one 'cube' place (ghana-pada) and two 'non- cube' places (aghana-pada). Then subtracting the (greatest possible) cube from the (last) 'cube' place and placing the (cube) root underneath the third place (to the right of the last 'cube' place), divide out the remainder up to one place less (than, that occupied by the cube root) by thrice the square of the cube root, which is not destroyed. Setting down the quotient (obtained from division) in the line (of the cube root), [and designating the quotient as the 'first' (ādima) and the cube root as the last' (antya)], subtract the square of that quotient, as multiplied by thrice the last' (antya), from one place less than that occupied by the quotient (uparima-rāśi) as before, and the cube of the 'first' (ādima) from its own place. (The number now standing in the line of cube root is the cube root of the given number up to its last-but-one cube place from the left). Again apply the rule, '(placing the cube root) under the third place' etc. (provided there be more than two 'cube' places in the given number; and continue the process till all cube places are exhausted). This will give the (cube) root (of the given number).

To illustrate this method, we find out the cube root of

Example: Cube root of 277167808

Denote the 'cube (c)' and 'non-cube (n)'places over the digits starting from the unit place

n	n	c	n	n	c	n	n	c
2	7	7	1	6	7	8	0	8
←

Subtract the greatest possible cube (6³ = 216) from the last cube place (277), 277 - 216 = 61. Write the cube root (6) underneath the third place to the right of the last 'cube' place

n	n	c	n	n	c	n	n	c
	6	1	1	6	7	8	0	8	← Remainder
				6					← Line of cube root

Dividing out by thrice the square of the cube root (i.e., by 3 x 6² = 108) the remainder up to one place less than that occupied by the cube root (i.e., 611). Here the quotient is 5 and the remainder is 611 - 540 =71. Write the quotient (5) in the line of the cube root (to the right of the cube root), we have

n	n	c	n	n	c	n	n	c
		7	1	6	7	8	0	8	← Remainder
				6	5				← Line of cube root

Let now the quotient 5 be called the 'first' and the cube root 6 the 'last'. Then subtracting the square of the 'first' as multiplied by thrice the 'last' (i.e., 3×6×5² = 450) from one place less than that occupied by the quotient (i.e., from 716), 716 - 450 = 266 we get

n	n	c	n	n	c	n	n	c
		2	6	6	7	8	0	8	← Remainder
				6	5				← Line of cube root

And subtracting the cube of the 'first' (i.e., 5³ = 125) from its own place (i.e. from 2667), 2667 - 125 = 2542 we get

n	n	c	n	n	c	n	n	c
		2	5	4	2	8	0	8	← Remainder
				6	5				← Line of cube root

One round of the operation is now over; and the number 65 standing in the line of the cube root is the cube root of the given number (277167808) up to its last-but-one 'cube' place from the left (i.e., of 277167).

As there is one more 'cube' place on the right, the process is repeated. Thus, placing the cube-root (i.e., 65) under the third place beginning with the last but one 'cube' place, we have

n	n	c	n	n	c	n	n	c
		2	5	4	2	8	0	8	← Remainder
						6	5		← Line of cube root

Dividing out 25428 by 3 x 65² = 12675 as before. Here the quotient is 2 remainder is 25428 - 25350 = 78 and placing the quotient (2) in the line of the cube root, we have

n	n	c	n	n	c	n	n	c
					7	8	0	8	← Remainder
						6	5	2	← Line of cube root

Here the quotient 2 is called the 'first' and the cube root 65 the 'last'. Then subtracting the square of the 'first' as multiplied by thrice the 'last' ( i.e. 3 x 2² x 65 = 780) from one place less than that occupied by the quotient (i.e., from 780), 780 - 780 = 0 we get

n	n	c	n	n	c	n	n	c
								8	← Remainder
						6	5	2	← Line of cube root

And finally subtracting the cube of the 'first' (i.e., 2³ = 8) from its own place (i.e. from 8), 8 - 8 = 0 we get

n	n	c	n	n	c	n	n	c
								0	← Remainder
						6	5	2	← Line of cube root

The second round of the operation is now completed. No more 'cube' places on the right, the process ends. The quantity in the line of the cube root, viz., 652, is the cube root of the given number. The remainder being zero, the cube root is exact.

Cube root of 277167808 = 652

Example: Cube root of 12812904

Denote the 'cube (c)' and 'non-cube (n)'places over the digits starting from the unit place

n	c	n	n	c	n	n	c
1	2	8	1	2	9	0	4
←

Subtract the greatest possible cube (2³ = 8) from the last cube place (12), 12 - 8 = 4. Write the cube root 2 underneath the third place to the right of the last 'cube' place

n	c	n	n	c	n	n	c
	4	8	1	2	9	0	4	← Remainder
			2					← Line of cube root

Dividing out by thrice the square of the cube root (i.e., by 3 x 2² = 12) the remainder up to one place less than that occupied by the cube root (i.e., 48). Here the quotient is 3 and the remainder is 48 - 36 =12. In order to continue with the rest of the steps, we have taken the quotient as 3 not 4 (12 x 4 = 48). Write the quotient (3) in the line of the cube root (to the right of the cube root), we have

n	c	n	n	c	n	n	c
	1	2	1	2	9	0	4	← Remainder
			2	3				← Line of cube root

Let now the quotient 3 be called the 'first' and the cube root 2 the 'last'. Then subtracting the square of the 'first' as multiplied by thrice the 'last' (i.e., 3×2×3² = 54) from one place less than that occupied by the quotient (i.e., from 121), 121 - 54 = 67 we get

n	c	n	n	c	n	n	c
		6	7	2	9	0	4	← Remainder
			2	3				← Line of cube root

And subtracting the cube of the 'first' (i.e., 3³ = 27) from its own place (i.e. from 672), 672 - 27 = 645 we get

n	c	n	n	c	n	n	c
		6	4	5	9	0	4	← Remainder
			2	3				← Line of cube root

One round of the operation is now over; and the number 23 standing in the line of the cube root is the cube root of the given number (12812904) up to its last-but-one 'cube' place from the left (i.e., of 12812).

As there is one more 'cube' place on the right, the process is repeated. Thus, placing the cube-root (i.e., 23) under the third place beginning with the last but one 'cube' place, we have

n	c	n	n	c	n	n	c
		6	4	5	9	0	4	← Remainder
					2	3		← Line of cube root

Dividing out 6459 by 3 x 23² = 1587 as before. Here the quotient is 4 remainder is 6459 - 6348 = 111 and placing the quotient (4) in the line of the cube root, we have

n	c	n	n	c	n	n	c
			1	1	1	0	4	← Remainder
					2	3	4	← Line of cube root

Here the quotient 4 is called the 'first' and the cube root 23 the 'last'. Then subtracting the square of the 'first' as multiplied by thrice the 'last' ( i.e. 3 x 23 x 4² = 1104) from one place less than that occupied by the quotient (i.e., from 1110), 1110 - 1104 = 6 we get

n	c	n	n	c	n	n	c
						6	4	← Remainder
					2	3	4	← Line of cube root

And finally subtracting the cube of the 'first' (i.e., 4³ = 64) from its own place (i.e. from 64), 64 - 64 = 0 we get

n	c	n	n	c	n	n	c
							0	← Remainder
					2	3	4	← Line of cube root

The second round of the operation is now completed. No more 'cube' places on the right, the process ends. The quantity in the line of the cube root, viz., 234, is the cube root of the given number. The remainder being zero, the cube root is exact.

Cube root of 12812904 = 234

See Also

पाटीगणितम् में 'घनमूल'