My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree3Thanks
  • 1 Post By skipjack
  • 2 Post By romsek
Reply
 
LinkBack Thread Tools Display Modes
January 10th, 2017, 11:09 AM   #1
Senior Member
 
Joined: Feb 2016
From: seattle

Posts: 371
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
File Type: png 5.PNG (3.6 KB, 12 views)

Last edited by skipjack; January 10th, 2017 at 11:15 AM.
GIjoefan1976 is offline  
 
January 10th, 2017, 11:21 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 18,250
Thanks: 1439

Hint: $1080x^7y^3 = 5x\left(6x^2y\right)^3$
Thanks from topsquark
skipjack is offline  
January 10th, 2017, 02:45 PM   #3
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 1,656
Thanks: 842

Quote:
Originally Posted by GIjoefan1976 View Post
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.

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

$\sqrt[3]{1080} = 6\sqrt[3]{5}$
Thanks from skeeter and topsquark
romsek is online now  
January 10th, 2017, 02:57 PM   #4
Senior Member
 
Joined: Feb 2016
From: seattle

Posts: 371
Thanks: 10

Quote:
Originally Posted by skipjack View Post
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.
GIjoefan1976 is offline  
January 10th, 2017, 03:15 PM   #5
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 1,656
Thanks: 842

Quote:
Originally Posted by GIjoefan1976 View Post
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.
romsek is online now  
January 10th, 2017, 03:26 PM   #6
Math Team
 
Joined: Jul 2011
From: Texas

Posts: 2,678
Thanks: 1339

Quote:
Originally Posted by romsek View Post

$1080 = 540 \cdot 2 = 270 \cdot 2^2 = 135 \cdot 2^3 = 27 \cdot 5 \cdot 2^3 = 3^3 \cdot 5 \cdot 2^3 = (2\cdot 3)^3 \cdot 5 = 6^3 \cdot 5$
as demonstrated above ... prime factor decomposition
skeeter is offline  
January 10th, 2017, 05:28 PM   #7
Senior Member
 
Joined: May 2016
From: USA

Posts: 857
Thanks: 348

Quote:
Originally Posted by GIjoefan1976 View Post
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.
JeffM1 is offline  
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.
deesuwalka is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

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 03:14 PM
Dividing Very Large Numbers nogar Elementary Math 1 March 14th, 2014 04:39 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 05:06 AM





Copyright © 2017 My Math Forum. All rights reserved.