But verbal expositions are not always so clear. The topic of this section is a rule for finding the "side," or cube root, of a perfect cube. The rule was inserted by a later hand as a kind of appendix to some manuscripts of the 10th-century *Epistles* of the Brethren of Purity (Ikhwa̅n a̅l-Ṣafa ͑). The *Epistles*, which were intended as a compendium of all scientific and philosophical knowledge of the time, were written by an anonymous, secretive group of scholars centered in Basra, in southern Iraq. Think of them as a kind of medieval version of the Bourbaki group.

Below is a translation of the appendix. Because the verbal explanation may be difficult to follow, we follow it with two worked-out examples.

Abu̅ Ṭa̅lib Aḥmad bin Ja ͑far bin Ḥamma̅d recounted that he found a characteristic property of the side of the cubic number that is being mentioned, which is that when asked for the side of a cubic number, the way [to find] it is that you take a sixth of the number being investigated. You see if the number [being investigated] is even, [in which case] you add up the squares of the odd numbers in a series until [you reach] what resulted from the sixth, and you multiply the remainder of the sum of the squares by six, to get the sought-after side. If the number being sought is odd, you take the squares of the even numbers in a series and you add them up. Where the sum ends, you look at the remainder and you multiply six by it, which gives the side. If the number that is investigated is odd, take the even squares in a series and add them up to whereof the addition ends, then look into the remainder and multiply it by six, wherein the side comes out.^{3}

We illustrate the rule by examples for both the even and the odd case.

**Example 1**

Suppose we wish to compute the cube root of \(14^3=2744\). Dividing \(2744\) by \(6\), we get \(457{1\over 3}\).

Then we add up the squares of the odd integers and we stop just before exceeding the latter number: \[1^2+3^2+5^2+7^2+9^2+11^2+13^2=455.\] The difference \(457{1\over 3}-455\) is \(2{1\over 3}\), and multiplying by \(6\) we find the cube root of \(2744\): \(6\times 2{1\over 3}=14\).

**Example 2**

We compute the cube root of \(17^3=4913\). Dividing \( 4913\) by \(6\) gives \(818{5\over 6}\).

Adding the squares of the even numbers as long as we stay below that gives \[2^2+4^2+\cdots+16^2=816.\] The difference is \(818{5\over 6}-816=2{5\over 6}\), and so multiplying by \(6\) we get \(17\).

In the Appendix, we provide more examples of Abu̅ Ṭa̅lib's rule in the form of a set of student exercises.

We have not seen this rule in any other text, either in Arabic or in any other language. Its obscurity can be accounted for by the fact that it is not at all a practical means of finding the cube root of a perfect cube, and it fails to give an approximation to the cube root of other numbers. (See the second student exercise in the Appendix.) It is a curiosity of number theory, never intended as an alternative to the standard rules practiced at the time, and equivalent to our Ruffini-Horner algorithm.^{4}

Abu̅ Ṭa̅lib's rule is stated with no hint of how it was derived. But its derivation is not difficult to figure out. First, note that \[n^3 - n = (n-1)n(n+1)\]

is the product of three consecutive integers. Al-Karaji̅, for one, stated and proved this identity in the early 11th century as a lemma to his rule for finding the sum \(1\cdot 2\cdot 3+2\cdot 3\cdot 4+3\cdot 4\cdot 5+\ldots+8\cdot 9\cdot 10\).^{5} Therefore, the cube root \(n\) of \(n^3\) can be expressed as the difference \[n = n^3 - (n-1)n(n+1).\]

Now, if we remember that the product of three consecutive numbers occurs in the rules for summing odd and even squares (provided we do not simplify the answer!), we obtain the method discovered by Abu̅ Ṭa̅lib. If \(n\) is even, then \(n-1\) is odd, and vice-versa, thus the requirement to add the odd squares for an even cube, and the even squares for an odd cube. For the case that \(n\) is even, we have \[n = n^3 - (n-1)n(n+1)=6\left({\rm\frac{1}{6}}n^3-(1^2+3^2+5^2+\ldots+(n-1)^2)\right),\]

while for \(n\) odd it is \[n = n^3 - (n-1)n(n+1)=6\left({\rm\frac{1}{6}}n^3-(2^2+4^2+6^2+\ldots+(n-1)^2)\right).\]

In an age when modern algebraic formulas for numerical identities were still in the distant future, mathematicians could naturally conceive of the sums of consecutive squares of odd numbers and of even numbers as a sixth of the upper extreme multiplied by the product of the next two integers. With this understanding of the rules, the discovery of Abu̅ Ṭa̅lib's method to calculate cube roots of perfect cubes, described above, becomes plausible and easy to understand. On the other hand, modern-day mathematicians, accustomed to the simplified formulas \[\sum_{i=0}^n (2i+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}\] and \[\sum_{i=1}^n (2i)^2 = \frac{2n(n+1)(2n+1)}{3}\] afforded by our algebraic notation, may have to work harder to understand why such a method works. This is in fact what happened to one of us, who understood the rule only after he re-wrote (1) and (3) in the non-simplified forms (2) and (4), and from this experience conceived the idea for writing this note.

##### Notes

3. Translation adjusted by the authors from [1, p. 68]. This Abu̅ Ṭa̅lib is otherwise unknown.

4. See [3] for such examples of cube root extraction from China, the Muslim world, and medieval Europe.

5. [4,141.1]