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There
is no evidence that the following method for solving the cube was known
earlier than Aryabhata (500 BC). It is stanza 5 of his Aryabhatiya that tells
of this method:
One should divide the second aghana by three times the square of the
cube roots of the preceeding ghana. The square (of the quotient) multiplied
by three times the purva (that part of the cube root already found) is to be
subtracted from the first aghana and the cube (of the quotient of the above
division) is to be subtracted from the ghana.
Certain
steps have been left out in Aryabhata's method for calculating the cube root.
This may have been due to limitations of the Sanscrit language. It was common
at this time to pass on teachings orally, hence, it is understandable that
some written methods may be vague.
Here
is an example of how Aryabhata solved the cube, taken from W.E. Clark:
Aryabhatiya (Chicago University Press), 1930.
Find
the cube of 1860867. Counting from right to left, the first, fourth, seventh
and so on places are named ghana(cubic), the second, fifth, eighth and so on
are called the first aghana (noncubic), while the third, sixth, ninth places
and so on are called the second aghana.
So in this example we start by taking the cube root of 1=1
1-1=0, and you bring down the 8.
08
Three times the square of the root= 3((1)2)=3
8=3(2)+2, so the first 2=the quotient or next digit of the root. So we have a
remainder of 2 and now we bring down the 6.
26
Square of the quotient multiplied by three times the purva=((2)2)(3)(1)=12.
26-12=14, now bring down
the 0
140
The cube of the quotient is (2)3=8 140-8=132,
now bring down the 8.
1328
Three times the square of the root=3((12)2)=432
1328= 432(3) + 32, so 3 is the quotient or next digit of the root. We
have a remainder of 32, and we bring down the 6.
326
Square of the quotient multiplied by 3 times the purva =((3)2)(3)(12)=324
326-324=2, bring down the 7.
27
cube of the quotient =(3)3=27 27-27=0,
therefore the cube root of 1860867 is 123.
Consider the problem of constructing a square twice the area of a given
square (A) of side 1 unit. It is clear that for the larger square (C) to have
twice the area of square A, it should have side  2 units. Also, we are given a third square
(B) of side 1 which needs to be dissected and reassembled so that by fitting
cut-up sections of square (C) on square (A), it is possible to make up a
square close to the size of square (C).
The
instructions in the Sulbasutras may be translated as: "
Increase the measure (ie. 1) by its third and this third by its own
fourth less the thirty-fourth part of that fourth. This is the value (of 2) with a special quantity in excess. "
A commentator on the Sulbasutras, Rama, who lived in the middle of the
fifteenth century gave an improved approximation by adding a fifth and sixth
term to the right-hand side of the equation, which then gave the first
seven places of decimals correct.
Bhaskaracharya (1114-1185)
Bhaskaracharya otherwise known as Bhaskara is probably the most well known
mathematician of ancient Indian today. Bhaskara was born in 1114 A.D. according
to a statement he recorded in one of his own works. He was from Bijjada Bida
near the Sahyadri mountains. Bijjada Bida is thought to be present day
Bijapur in Mysore state.
Bhaskara wrote his famous Siddhanta Siroman in the year 1150 A.D. It is
divided into four parts; Lilavati (arithmetic), Bijaganita (algebra),
Goladhyaya (celestial globe), and Grahaganita (mathematics of the planets). Much
of Bhaskara's work in the Lilavati and Bijaganita was derived from earlier
mathematicians; hence it is not surprising that Bhaskara is best in dealing
with../5thCenturyBC/ModKuttaka indeterminate analysis. In connection with the
Pell equation, x2=1+61y2, nearly solved by
7thCentury BC Brahmagupta , Bhaskara gave a method - Chakravala process for
solving the equation.
Bhaskara was somewhat of a poet as were many Indian mathematicians at this
time. Here are a couple of lines found in Lilavati that demonstrate his
poetic inclination:
"O girl! out of a group of swans, 7/2 times the square root of the
number are playing on the shore of a tank. The two remaining ones are playing
with amorous fight, in the water. What is the total number of swans?"
Teaching and learning mathematics was in Bhaskara's blood. He learnt
mathematics from his father, a mathematician, and he himself passed his
knowledge to his son Loksamudra.
Cubic
equation.
Examples
found in the "Siddhanta_Siroman"
Here is an example of how a cubic equation was solved in "Bhaskara's
Bijaganita" (1150) (This example was used in C.N.
Srinivasiengar's: The History of Ancient Indian Mathematics.)
The
Bijaganita gives one example of his solution (completing the cube) of this
special cubic equation.
x3 + 12x = 6x2 + 35.
Bhaskara
rewrites this equation as;
x3 - 6x2 + 12x - 8 = 27. And
therefore, (x - 2)3 = 33.
Taking
the cube root of both sides we get x - 2 = 3. Thus, x = 5.
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