User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 October 2nd, 2011, 08:03 AM #1 Newbie   Joined: Oct 2011 Posts: 10 Thanks: 0 Solutions of the equation 1. Determine all the solutions of (x^2+3x-4)^3 + (2x^2-5x+3)^3 + (3x^2-2x-1)^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, 08:16 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, 08:26 AM   #3
Newbie

Joined: Oct 2011

Posts: 10
Thanks: 0

Re: Solutions of the equation

Quote:
 Originally Posted by mattpi 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
By doing as you said, I get (x-1)^3 as common for (x+4)^3 + (2x-3)^3 + (3x+1)^3.
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, 09:31 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, 10:35 AM   #5
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

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:
 [color=beige]. . [/color]

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, 03:41 AM #6 Newbie   Joined: Oct 2011 Posts: 10 Thanks: 0 Re: Solutions of the equation Thanks @soroban. Nice solution!!  Tags equation, solutions how many solutions are there to the equation calculator

Click on a term to search for related topics.
 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 raycol1970 Algebra 3 April 14th, 2012 02:03 PM Solarmew Applied Math 12 November 21st, 2011 05:01 PM jakeward123 Calculus 5 June 3rd, 2011 12:17 AM rebecca Calculus 3 July 31st, 2010 05:35 AM raycol1970 Applied Math 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      