
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
January 10th, 2017, 11:09 AM  #1 
Senior Member Joined: Feb 2016 From: seattle Posts: 354 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. Last edited by skipjack; January 10th, 2017 at 11:15 AM. 
January 10th, 2017, 11:21 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 17,466 Thanks: 1312 
Hint: $1080x^7y^3 = 5x\left(6x^2y\right)^3$

January 10th, 2017, 02:45 PM  #3  
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650  Quote:
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, 02:57 PM  #4 
Senior Member Joined: Feb 2016 From: seattle Posts: 354 Thanks: 10  
January 10th, 2017, 03:15 PM  #5  
Senior Member Joined: Sep 2015 From: CA Posts: 1,264 Thanks: 650  Quote:
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: 2,569 Thanks: 1272  Quote:
 
January 10th, 2017, 05:28 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 678 Thanks: 283  Quote:
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, 01:51 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 01:55 AM. 

Tags 
cube, finding, large, numbers 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Are large numbers definable?  jk12  Math  42  May 18th, 2015 02:14 PM 
Dividing Very Large Numbers  nogar  Elementary Math  1  March 14th, 2014 03:39 PM 
The last two digits of LARGE numbers?  ricsi046  Number Theory  2  November 10th, 2013 06:31 AM 
Weak Law of Large Numbers  Artus  Advanced Statistics  0  January 29th, 2013 01:14 AM 
Finding percentages of large numbers  Kitty  Elementary Math  1  October 28th, 2008 04:06 AM 