पाटीगणित (श्रीधराचार्य - प्राचीन संस्कृत टीका एवं प्रो. कृपाशंकर शुक्ल सविस्तार व्याख्या)

12 LOGISTICS
Adding up these numbers, we get 15625. This is the cube of 25.
Now we treat 25 as the last and 6 as the first, and proceeding as
above we get 16777216 as the cube of 256.
The second and third rules are :
(i) 𝑛³ = 𝑛 × 𝑛 × 𝑛
(ii) 𝑛³ = (𝑛 - 1)³ + 3𝑛 (𝑛 - 1) + 1 .
Datta and Singh, assuming
खैकादिचये वाऽन्त्ये त्र्यादिहते पूर्वघनयुतिः सैके ॥
to be the reading of the second half of verse 28, have given the following
translation for it : ¹
"The cube (of a given number) is equal to the series whose terms
are formed by applying the rule, 'the last term multiplied by thrice the
preceding term plus one,' to the terms of the series whose first term is zero,
'the common difference is one and the last term is the given number."
This gives
𝑛³ = Σ [ 3𝑟(𝑟 - 1) + 1 ]
in place of formula (ii) above.²
Ex. 5. Quickly say what are the cubes of 1 to 9, 15,
256, and 203.
(8) Cube root (ghana-mûla)
Rule for finding the cube root :
29-31. (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 pos-
sible) 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' (âdima) 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-râśi) as before, and

¹ B. Datta and A. N. Singh, History of Hindu Mathematics, Part I,
page 168.
² This formula is also given in GSS, ii. 45.