## Root extraction, part II: cube roots

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As you might guess, this post builds on “Root extraction, part I“, which gave a way to visualize the traditional square root algorithm geometrically, an approach that has the advantage that each step appears natural and easily motivated.

Our goal herein is to do much the same for cube roots. The point is to find a geometric construction, ideally one well-suited to physical manipulatives, in which the steps in building the successive digits of the cube root of a number are transparent.  As with the post on square roots, I make no claims to originality in what follows.

Example:  Find $\sqrt[3]{22665187}$

Oh, nice.

We’re looking for a cube whose volume is 22665187.  Comparing 100³=1,000,000 with 1000³=1,000,000,000, we see our answer is in the hundreds; further, the cube root lies between 200 and 300, since 200³=8,000,000 < 22,665,187 < 300³=27,000,000.

We know the hundreds digit of our root is “2”; we create a cube, where each edge has its length split into two segments, one of length 200, the other of length 10d, where “d” is the tens digit of the root.  Our goal is to determine an appropriate choice for d so that the volume of this cube is close to but no more than 22,665,187.

In the diagram above, the red core cube is (200)(200)(200); each of the three green faces has dimensions (200)(200)(10d); each of the three yellow seams has dimensions (200)(10d)(10d); the final little cube has dimensions (10d)(10d)(10d).  The outer fringes of the big cube (the non-red portions) are meant to have volume nearly equal to 22,665,187 – 8,000,000= 14,665,187.  But since d is small, we expect the yellow and blue volumes to be negligible relative to the green faces, and so we compare 3x400000d to 14,665,187, which leads to a guess of d = 12.  (!)  Apparently, those negligible seams are bigger than we thought!

If we guess d=9, we find the volume of the big cube to be 8,000,000 [red]+ 3x(40000)(90) [green] + 3x(200)(90)(90)[yellow] + (90)(90)(90)[blue] = 24,389,000, a bit too big.

Revising our guess to be d=8, we find the volume of the big cube to be 8,000,000 [red]+ 3x(40000)(80) [green] + 3x(200)(80)(80)[yellow] + (80)(80)(80)[blue]=21,952,000, a bit smaller than our target.  Thus we’ve determined that the tens place of our cube root is an “8”.

We repeat the process, now with a cube whose sides have length 280+d, and where d is the ones place of our root.

The red inner cube has volume 280³=21,952,000 from our earlier work, and so the remaining regions seek to have volume 713,817.  As before, we’ll focus on only the three green regions initially, as the others are quite small by comparison.  The green flats each have dimensions 280 by 280 by d, and so the total volume is (3)(280)²(d)=235200d.  Comparing 235,200d to 713817, we find that d can be no more than 3.  Setting d=3, we find a total volume of the big cube of 280³ [red] + (3)(280)²(3) [green]+3(280)(3)²[yellow]+3³[blue], which yields 22,665,187.

Voila!  We’ve determined that $\sqrt[3]{22665187} = 283$.

Now it is indeed possible to apply this very same technique to finding cube roots of any number to an arbitrary number of decimal places; one is only constrained by one’s willingness to compute the volumes of the green faces (which forces you to triple the square of the digits found so far for the root), the volumes of the yellow strips (multiplying the digits found so far for the root by 3 x the next digit squared), and the volume of the blue cube (the next digit cubed).    Naturally, this is precisely the traditional cube root algorithm (and precisely the ancient Chinese algorithm for root extraction with number rods or with an abacus).

I gather that classroom manipulatives were in vogue at the beginning of the 20th century, consisting essentially of a large block, three flats, three strips, and a small block (much as in our diagram above); the intent was that pupils would use the manipulative as a support as they extracted cube roots by hand of various numbers.

If one wished to extend this idea to higher order roots, one could, but one loses the ability to do the calculation geometrically, at least with a tangible object; the binomial theorem and algebraic reasoning become necessary tools for computing fifth and higher order roots by these means.

### 7 Responses to “Root extraction, part II: cube roots”

1. jd2718 Says:

Hi! Thanks for this.

Where do the diagrams come from? (very nice visual for centered hexagonal numbers, believe it or not)

Jonathan

2. TwoPi Says:

I built them in Microsoft Word, trying to capture the essence what I had done in class using base ten blocks.

3. jd2718 Says:

very nice. I know it’s not your problem, but look how easy it is to see that $(n+1)^3 - n^3 = 3(n-1)^2 + 3(n-1) + 1$

Oh, yeah, the post is impressive, too. I’m just stuck on that diagram.

4. Carnival of Mathematics 1000 « JD2718 Says:

[…] Borovik discusses at Math Under the Microscope. 00 – A geometric interpretation of how to extract cube roots (for the brave) over at Blog 360. (Also, for the brave and non-brave alike, discussion of how to […]

5. Michael Says:

You might be interested to know that this basic approach was actually used in schools in the late 19th and early 20th centuries. Three dimensional wooden models called Cube Root Blocks were available for teachers to use as visualization aids. You can find an example of a “Double” cube root block on my website at sawbonesantiques.com under the “Technical” link. The double block allowed calculation of roots to three significant digits.

6. jovelyn Says:

hey! thank you verry much for the illustration and discussion on how to extract the cube root of a number……………..

7. Auroran Says:

many higher order roots can be obtained by multiplying by other roots, the difficulty seems to be in dealing with…..prime roots-excepting the number 3, of course. 🙂