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This is based on a lesson at Khan Academy that I didn't understand.

In the lesson, the instructor uses the number 512 as an example and the entire prime factorization consists of three groups of three 2s. However, not every number has such a nice prime factorization and I don't understand how to use prime factorization to determine a cube root.

For example, if the number is 1000, I know the cube root is 10, but how do I get there using prime factorization.

The prime factorization of 1000 is 2 * 2 * 2 * 5 * 5 * 5. Since it's six prime numbers, I can't get 3 equal groups of the same number like I could with 512. In another forum someone said to use three groups of the smallest factor, but three 2s doesn't get me 10.

What am I missing?

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  • $\begingroup$ You're missing a copy of $2$. $1000 = 2^{3}5^{3}$. $\endgroup$
    – Elchanan Solomon
    Commented Apr 23, 2013 at 19:16

3 Answers 3

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You're missing a third factor of $2$ in your prime factorization of $1000$:

$$1000 = 2\cdot 2 \cdot 2 \cdot 5 \cdot 5 \cdot 5 = (2^3\cdot 5^3) = (2\cdot 5)^3 = (2\times 5) \cdot (2\times 5) \cdot (2 \times 5) = 10 ^3$$

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  • $\begingroup$ Can you see how to reorder the terms, to get three groups/factors of $10$? $\endgroup$
    – amWhy
    Commented Apr 23, 2013 at 19:23
  • $\begingroup$ We simply rearrange the order of the factors (which doesn't change their product): $2\cdot 2\cdot 2 \cdot 5 \cdot 5\cdot 5 = 2 \cdot 5 \cdot 2 \cdot 5 \cdot 2 \cdot 5 = 10 \cdot 10 \cdot 10$ $\endgroup$
    – amWhy
    Commented Apr 23, 2013 at 19:25
  • $\begingroup$ I think I get it. I was thinking I had to use the numbers in order. I forgot that the order is arbitrary. DOH! So, my three groups are all 2*5. Now that I realize this, I think I can apply it to other numbers. Thanks! $\endgroup$
    – user74014
    Commented Apr 23, 2013 at 19:27
  • $\begingroup$ You're welcome! Yes, multiplication is commutative on the integers: so $m\times n = n\times m$, etc. $\endgroup$
    – amWhy
    Commented Apr 23, 2013 at 19:31
  • $\begingroup$ @amWhy: you do such a great job of following up! +1 $\endgroup$
    – Amzoti
    Commented Apr 24, 2013 at 0:35
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The prime factorization for 1000 is 2*2*2*5*5*5. So you have three groups you're trying to find.

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  • $\begingroup$ You're just missing another 2 in your prime factorization. So it looks like you're probably doing everything else correctly in finding the 3 groups of numbers, etc. $\endgroup$
    – agktmte
    Commented Apr 23, 2013 at 19:17
  • $\begingroup$ I corrected my error, but I don't get 10 with three 2s or with three 5s, so I still don't get it. $\endgroup$
    – user74014
    Commented Apr 23, 2013 at 19:21
  • $\begingroup$ Combine the groups: 2*2*2*5*5*5 = 2*5 * 2*5 * 2*5 if you reorder them, so it becomes 10*10*10, just rearrange the prime factors. $\endgroup$
    – agktmte
    Commented Apr 23, 2013 at 19:22
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What you're missing is there are indeed 3 factors of 2 in 1000. Remember, when you multiply a number by 10, you add a 0 to the end of it, so it should be easy to see that

$$1000=10\times10\times10$$

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  • $\begingroup$ I corrected my error, but I don't get 10 from three 2s. I realize it's easy to memorize the cube root of a number like 1000. That was just an example. The question was how do I use prime factorization to get a cube root? $\endgroup$
    – user74014
    Commented Apr 23, 2013 at 19:23
  • $\begingroup$ @Ghodmode It looks like amWhy has that part covered. $\endgroup$
    – Mike
    Commented Apr 23, 2013 at 19:27

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