Algebra Pre-Algebra and Basic Algebra Math Forum

January 10th, 2017, 11:09 AM   #1
Senior Member

Joined: Feb 2016
From: seattle

Posts: 377
Thanks: 10

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.
Attached Images 5.PNG (3.6 KB, 12 views)

Last edited by skipjack; January 10th, 2017 at 11:15 AM. January 10th, 2017, 11:21 AM #2 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Hint: $1080x^7y^3 = 5x\left(6x^2y\right)^3$ Thanks from topsquark January 10th, 2017, 02:45 PM   #3
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,638
Thanks: 1475

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 $\sqrt{1080}$

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

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

$\sqrt{1080} = 6\sqrt{5}$ January 10th, 2017, 02:57 PM   #4
Senior Member

Joined: Feb 2016
From: seattle

Posts: 377
Thanks: 10

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, 03:15 PM   #5
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,638
Thanks: 1475

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, 03:26 PM   #6
Math Team

Joined: Jul 2011
From: Texas

Posts: 3,094
Thanks: 1677

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, 05:28 PM   #7
Senior Member

Joined: May 2016
From: USA

Posts: 1,310
Thanks: 552

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.

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

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

I am not sure that process is enough more efficient than just running through the primes to warrant memorization. January 11th, 2017, 01:51 AM #8 Member   Joined: Sep 2016 From: India Posts: 88 Thanks: 30 $\sqrt{1080x^7y^3}$ $=\sqrt{216×5×x^3×x^3×x×y^3}$ $=x^2\cdot y\sqrt{6^3×5x}$ $=x^2\cdot y\cdot6\sqrt{5x}$ Last edited by deesuwalka; January 11th, 2017 at 01:55 AM. Tags cube, finding, large, numbers Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jk12 Math 42 May 18th, 2015 03:14 PM nogar Elementary Math 1 March 14th, 2014 04:39 PM ricsi046 Number Theory 2 November 10th, 2013 06:31 AM Artus Advanced Statistics 0 January 29th, 2013 01:14 AM Kitty Elementary Math 1 October 28th, 2008 05:06 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      