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January 10th, 2017, 10:09 AM   #1
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finding the cube of large numbers?

Is there a way to figure out these larger numbers I get how to do the variables though so far.

Thanks.
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Last edited by skipjack; January 10th, 2017 at 10:15 AM.

 January 10th, 2017, 10:21 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,826 Thanks: 2160 Hint: $1080x^7y^3 = 5x\left(6x^2y\right)^3$ Thanks from topsquark
January 10th, 2017, 01:45 PM   #3
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Quote:
 Originally Posted by GIjoefan1976 Is there a way to figure out these larger numbers I get how to do the variables though so far. Thanks.
I'm going to assume the main problem here is to find $\sqrt[3]{1080}$

The only sure fire way I know of is to completely factor $1080$ and pull the perfect cubes out of it.

$1080 = 540 \cdot 2 = 270 \cdot 2^2 = 135 \cdot 2^3 = 27 \cdot 5 \cdot 2^3 = 3^3 \cdot 5 \cdot 2^3 = (2\cdot 3)^3 \cdot 5 = 6^3 \cdot 5$

$\sqrt[3]{1080} = 6\sqrt[3]{5}$

January 10th, 2017, 01:57 PM   #4
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Quote:
 Originally Posted by skipjack Hint: $1080x^7y^3 = 5x\left(6x^2y\right)^3$
Thanks for the hint, but guessing there is no shortcut or "trick" I saw something about trying to memorize the cubed roots of 1 -10 but that looks like a lot of work for me right now.

January 10th, 2017, 02:15 PM   #5
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Quote:
 Originally Posted by GIjoefan1976 Thanks for the hint, but guessing there is no shortcut or "trick" I saw something about trying to memorize the cubed roots of 1 -10 but that looks like a lot of work for me right now.
I can whip off 1 8 27

I have to think for a sec that $4^3 = 64$

I can whip off $5^3 = 125$

that's it other than $10^3=1000$

the rest I'd have to figure out.

So memorizing cubes isn't going to help that much.

January 10th, 2017, 02:26 PM   #6
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Quote:
 Originally Posted by romsek $1080 = 540 \cdot 2 = 270 \cdot 2^2 = 135 \cdot 2^3 = 27 \cdot 5 \cdot 2^3 = 3^3 \cdot 5 \cdot 2^3 = (2\cdot 3)^3 \cdot 5 = 6^3 \cdot 5$
as demonstrated above ... prime factor decomposition

January 10th, 2017, 04:28 PM   #7
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Quote:
 Originally Posted by GIjoefan1976 Thanks for the hint, but guessing there is no shortcut or "trick" I saw something about trying to memorize the cubed roots of 1 -10 but that looks like a lot of work for me right now.
You mean memorizing the cubes of the integers 1 through 10.

It seems like a relatively useless thing to memorize. It might POSSIBLY be worthwhile to memorize the cubes of 2, 3, 4, and 5 and the squares of 2 through 10 to reduce relatively simple radicals.

We have the cube root of an EVEN integer. Given that

$10^3 = 1000$ we are looking for cubes of primes (if any apply) that are small. 2 is small, and its cube is 8.

$\dfrac{1080}{8} = 135 \implies \dfrac{135}{5} = 27 = 3^3 \implies 1080 = 2^3 * 3^3 * 5 \implies$

$\sqrt[3]{1080} = \sqrt[3]{2^3 * 3^3 * 5} = 2 * 3 \sqrt[3]{5} = 6\sqrt[3]{5}.$

I am not sure that process is enough more efficient than just running through the primes to warrant memorization.

 January 11th, 2017, 12:51 AM #8 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $\sqrt[3]{1080x^7y^3}$ $=\sqrt[3]{216×5×x^3×x^3×x×y^3}$ $=x^2\cdot y\sqrt[3]{6^3×5x}$ $=x^2\cdot y\cdot6\sqrt[3]{5x}$ Last edited by deesuwalka; January 11th, 2017 at 12:55 AM.

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