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

CUBE ROOT 13
the cube of the ‘first’ (âdima) 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).¹
To illustrate this method, we find out the cube root of
277167808.
Indicating the ‘cube’ and ‘non-cube’ places by writing the tachy-
graphic abbreviations c and n over the digits, we have
n n c n n c n n c
2 7 7 1 6 7 8 0 8
Subtracting the greatest possible cube (i.e., 6³ or 216) from the last
‘cube’ place (i e., from 277) and placing the cube root (i.e., 6) underneath
the third place to the right of the last ‘cube’ place, we have
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 3.6² or 108)
the remainder up to one place less than that occupied by the cube root
(i.e., 611), and setting down the quotient 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, or 450) from one place less than that occupied by
the quotient (i.e., from 716), 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)

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¹ Cf. Â, ii. 5 ; BrSpSi, xii. 7 ; GSS, ii. 53, 54 ; MSi, xv. 9-10(i) ;
GT, p. 13, lines 18·25 ; Siśe, xiii. 6-7 ; L (ASS), vv. 28-29 ; GK, I, pp. 8-9,
vv. 24-25.