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

14 LOGISTICS
And subtracting the cube of the 'first' (i.e., 5³ or 125) from its own place
(i.e. from 2667), 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 stand-
ing in the line of the cube root is the cube root of the given number
(277167808) upto 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 3×65² (=12675) as before, and placing the
quotient 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)
Subtracting 3×65×2² (=780) from 780 we get
n n c n n c n n c
8 (remainder)
6 5 2 (line of cube root)
Finally subtracting 2³ from 8, 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 over. There being 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.