## On the solution to a cubic equation

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 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 GH be equal to 20. I 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 AB is equal to GH and therefore to the value of the ‘thing,’ for it was supposed of GH that it was so [i.e. equal to x],
therefore I complete, after the manner of the first theorem of the 6th chapter of this book, the solids DA, DC, DE, DF, so that we understand by DC the cube of BC, by DF the cube of AB, by DA three times CB times the square of AB, by DE three times AB times the square of BC. 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 GB 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 has been said in the 35th [of Book I] and 31st of the 11th Book of [Euclid’s] *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 have the

## Rule

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 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 \sqrt{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 ‘things’, \sqrt[3]{\sqrt{108} + 10} - \sqrt[3]{\sqrt{108} - 10}.

## Reference

- J. Fauvel, J. Grey,
*The History of Mathematics: A Reader*, The Open University, 1987, pp. 260-262