Cube in Pāṭīgaṇitam

Here we will be knowing how to find cube of a number as per Pāṭīgaṇitam.

Verse

स्थाप्योऽन्त्यघनोऽन्त्यकृतिः स्थानाधिक्यं त्रिपूर्वगुणिता च ।

आद्यकृतिरन्त्यगुणिता त्रिगुणा च घनस्तथाऽऽद्यस्य ॥ २७ ॥

निर्युक्तराशिरन्त्यं (तस्य) घनोऽसौ समत्रिराशिहतिः ।

एकादिचये वाऽन्त्ये त्र्यादिहते पूर्वघनयुतिः सैके ॥ २८ ॥

Translation

[Let the last digit of the given number be called the 'last' (antya) and the last-but-one digit the 'first' (ādi) or the 'preceding' (pūrva).]

Set down the cube of the 'last'; then set down, (successively) one place forward (sthānādhikyāṃ), (i) the square of the 'last' as multiplied by thrice the 'preceding', (ii) the square of the 'first' as multiplied by the 'last' as well as by 3, and (iii) the cube of the 'first'. This gives the cube of the combined number (formed by the 'last' and the 'first') (niryukta- rāśi), which should now be treated as the 'last' (provided there be more than two digits in the given number).

The (continued) product of three equal quantities; or, in the series having 1 for the first term and common difference (considering the last two terms, designating them as the 'first' or 'preceding' and the 'last' respectively), the last multiplied. by thrice the 'first', and increased by 1, and that added to the cube of the 'preceding', is also the cube.

Of the three rules stated here, the first one is the main method of cubing a number, To illustrate it by an example, let us find the

cube of 256.

Example: Cube of 256

To start with we find the cube of 25 with 2 as 'last' and 5 as 'first'.

last	first
2	5

Cube of the last (23)	8
Square of the 'last' as multiplied by thrice the 'first'. (3 x 5 x 22 = 60) to be placed in the next place such that 0 is one place next to 8	6	0
Square of the 'first' as multiplied by thrice the 'last'. (3 x 2 x 52 = 150) to be placed in the next place such that 0 is one place next to 0	1	5	0
Cube of the first (53 = 125) to be placed in the next place such that 5 is one place next to 0		1	2	5
Sum of the each columns = Cube of 25 =	15	6	2	5

Now we find the cube of 256 with 25 as 'last' and 6 as 'first'.

last	first
25	6

Cube of the last (253)	1	5	6	2	5
Square of the 'last' as multiplied by thrice the 'first'. (3 x 6 x 252 = 11250) to be placed in the next place such that 0 is one place next to 5		1	1	2	5	0
Square of the 'first' as multiplied by thrice the 'last'. (3 x 25 x 62 = 2700) to be placed in the next place such that 0 is one place next to 0				2	7	0	0
Cube of the first (63 = 216) to be placed in the next place such that 6 is one place next to 0						2	1	6
Sum of the each columns = Cube of 256 =	1	6	7	7	7	2	1	6

Cube of 256 = 16777216

See Also

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