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October 2nd, 2011, 09:03 AM  #1 
Newbie Joined: Oct 2011 Posts: 10 Thanks: 0  Solutions of the equation
1. Determine all the solutions of (x^2+3x4)^3 + (2x^25x+3)^3 + (3x^22x1)^3. Pretty tough one...I can't get it..Pls help.. Also try this one viewtopic.php?f=13&t=23598...Thnx.. 
October 2nd, 2011, 09:16 AM  #2 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: Solutions of the equation
Try factorising each of the cubed quadratic terms into the product of two cubed linear terms  i.e. write each term in the form  and see if that helps you at all

October 2nd, 2011, 09:26 AM  #3  
Newbie Joined: Oct 2011 Posts: 10 Thanks: 0  Re: Solutions of the equation Quote:
How to proceed afterwards???? Shall I open all the brackets or is there any one to simplify first? Also as you see it has come to a factorised form, are there only two factors or will there be more?  
October 2nd, 2011, 10:31 AM  #4 
Senior Member Joined: May 2008 From: York, UK Posts: 1,300 Thanks: 0  Re: Solutions of the equation
Once you've got the factor of I would suggest expanding the rest into one cubic and using either trial division or the rational root theorem to try and find the other roots.

October 2nd, 2011, 11:35 AM  #5  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 407  Re: Solutions of the equation Hello, mathswizard! You didn't give us an equation. I'll assume that the polynomials is set equal to zero. Quote:
Note that all the terms can be factored: [color=beige]. . . .[/color] [color=beige]. . [/color] [color=beige]. . . . . . . . . . .[/color] [color=beige]. . . . . . . [/color] [color=beige]. . . . . . . . . . . . . . . . . . .[/color] [color=beige]. . . . . . . . . . . . . . . . . . .[/color] [color=beige]. . . . . . . . . . . . . . . . . . . . . . . . . . .[/color]  
October 3rd, 2011, 04:41 AM  #6 
Newbie Joined: Oct 2011 Posts: 10 Thanks: 0  Re: Solutions of the equation
Thanks @soroban. Nice solution!! 

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