Concerning a Cube and "Things" Equal to a Number

Chapter XI

Scipio del Ferro of Bologna about thirty years ago invented [the method set forth in) this chapter, [and] communicated it to Antonio Maria Florido of Venice, who when he once engaged in a contest with Nicolo Tartalea of Brescia announced that Nicolo also invented it; and he [Nicolo] communicated it to us when we asked for it, but suppressed the demonstration. With this aid we sought the demonstration, and found it, though with great difficulty, in the manner which we set out in the following.

*Demonstration*

For example, let the cube of *GH *and six times the
side
*OH be *equal *to *20. 1 take two cubes *AE *and *CL
*whose
difference shall be 20, so that the product of the side *AC *by the
side *CK *shall be *2, -i.e., *a third of the number of "things;"
and I lay off *CB *equal to *CK, _{ }*then I say that
if it is done thus, the remaining line

Since therefore from *AC *times *CK *the result
is 2, from 3 times
*AC *times *CK *will result 6, the number
of "things;" and therefore from *AB *times 3 *AC *times *CK
*there
results 6 "things"
*AB, *or 6 times *AB, *so that 3 times the
product of *AB, BC,
*and *AC *is 6 times *AB.*

But the difference of the cube *AC *from the cube
*CK,
*and
likewise from the cube *BC, *equal to it by hypothesis, is 20; and
from the first theorem of the 6th chapter, this is the sum of the solids
*DA, DE, *and *DF, *so that these three solids make 20.

But taking *BC minus, *the cube of *AB *is equal
to the cube of
*AC *and 3 times *AC *into the square of *CB
*and
minus the cube of *BC *and minus 3 times *BC *into the square
of *AC.*

By the demonstration, the difference between 3 times *CB
*times
the square of *AC, *and 3 times *AC *times the square of *BC,
*is
[3 times] the product of *AB, BC, *and *AC.*

Therefore since this, as has been shown, is equal to 6
times *AB, *adding 6 times *AB *to that which results from *AC
*into
3 times the square of *BC *there results 3 times *BC *times the
square of *AC,
*since *BC *is minus.

Now it has been shown that the product of *CB *into
3 times the square of *AC *is minus; and the remainder which is equal
to that is plus, hence 3 times *CB *into the square of *AC *and
3 times *AC *into the square of *CB *and 6 times *AB *make
nothing.

Accordingly, by common sense, the difference between the
cubes *AC *and
*BC *is as much as the totality of the cube of
*AC,
*and
3 times
*AC *into the square of *CB, *and 3 times
*CB
*into
the square of *AC *(minus), and the cube of *BC *(minus), and
6 times *AB.*

This therefore is 20, since the difference of the cubes
*AC
*and
*CB
*was 20.

Moreover, by the second theorem of the 6th chapter, putting
*BC
*minus,
the cube of *AB *will be equal to the cube of *AC
*and 3 times
*AC
*into the square of *BC *minus the cube of
*BC
*and minus
3 times *BC *into the square of *AC.*

Therefore the cube of *AB, *with 6 times *AB, *by
common sense, since it is equal to the cube of *AC *and 3 times *AC
*into
the square of *CB, *and minus 3 times *CB *into the square of
*AC,*
and minus the cube of *CB *and 6 times *AB,
*which is now equal
to 20, as has been shown, will also be equal to 20.

Since therefore the cube of *AB *and 6 times *AB
*will
equal 20, and the cube of *GH, *together with 6 times *GH,
*will
equal 20, by common sense and from what has been said in the 35th and 31st
of the 11th Book of the *Elements, GH *will be equal to *AB,
*therefore
*GH
*is
the difference of *AC *and *CB. *But
*AC *and
*CB,
*or
*AC *and *CK, *are numbers or lines containing an area equal
to a third part of the number of "things" whose cubes differ by the number
in the equation, wherefore we have the

Cube the third part of the number of "things," to which you add the square of half the number of the equation,and take the root of the whole, that is, the square root, which you will use, in the one case adding the half of the number which you just multiplied by itself, in the other case subtracting the same half, and you will have a "binomial" and "apotome" respectively;

then subtract the cube root of the apotome from the cube root of the binomial, and the remainder from this is the value of the "thing."

In the example, the cube and 6 "things" equals 20; raise 2, the 3rd part of 6, to the cube, that makes 8; multiply 10, half the number, by itself, that makes 100; add 100 and 8, that makes 108; take the root, which is Square root of 108, and use this, in the first place adding 10, half the number, and in the second place subtracting the same amount, and you will have the binomial sqrt(108) + 10, and the apotome sqrt(108) - 10;take the cube root of these and subtract that of the apotome from that of the binomial, and you will have the value of the "thing,"

(cube root of (square root of 108)+ 10) - (cube root of (square root of 108) - 10).