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 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 $(ax^2+bx+c)^3$ term in the form $(\alpha_1x+\beta_1)^3(\alpha_2x+\beta_2)^3$ - and see if that helps you at all
October 2nd, 2011, 08:26 AM   #3
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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 $(ax^2+bx+c)^3$ term in the form $(\alpha_1x+\beta_1)^3(\alpha_2x+\beta_2)^3$ - 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 $(x-1)^3,$ 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
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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:
 $\text{1. Determine all the solutions of:}$ [color=beige]. . [/color]$(x^2\,+\,3x\,-\,4)^3\,+\,(2x^2\,-\,5x\,+\,3)^3\,+\,(3x^2\,-\,2x\,-\,1)^3 \;=\;0$

Note that all the terms can be factored:

[color=beige]. . . .[/color]$\bigg[(x-1)(x+4)\bigg]^3 + \bigg[(x-1)(2x-3)\bigg]^3 + \bigg[(x-1)(3x+1)\bigg]^3 \;=\;0$

[color=beige]. . [/color]$(x-1)^3(x+4)^3\,+\,(x-1)^3(2x-3)^3\,+\,(x-1)^3(3x+1)^3\;=\;0$

[color=beige]. . . . . . . . . . .[/color]$(x-1)^3\bigg[(x+4)^3\,+\,(2x-3)^3\,+\,(3x+1)^3\bigg]\;=\;0$

$\text{Note that: }\:(x+4)^3\,+\,(2x-3)^3\,\text{ is a sum of cubes.}$

$\text{It factors: }\:[(x+4)\,+\,(2x-3)]\,\left[(x+4)^2\,-\,(x+4)(2x-3) + (2x-3)^2\right]$
[color=beige]. . . . . . . [/color]$=\;(3x+1)(3x^2\,-\,9x\,+\,37)$

$\text{The equation becomes: }\:(x-1)^3\bigg[(3x+1)(3x^2-9x+37)\,+\,(3x+1)^3\bigg] \;=\;0$

[color=beige]. . . . . . . . . . . . . . . . . . .[/color]$(x-1)^3(3x+1)\bigg[(3x^2-9x+37)\,+\,(3x+1)^2\bigg] \;=\;0$

[color=beige]. . . . . . . . . . . . . . . . . . .[/color]$(x-1)^3(3x+1)\bigg[3x^2-9x+37+9x^2+6x+1\bigg] \;=\;0$

[color=beige]. . . . . . . . . . . . . . . . . . . . . . . . . . .[/color]$(x-1)^3(3x+1)\underbrace{(12x^2-3x+38)}_{\text{No real roots}} \;=\;0$

$\text{Therefore: }\;\begin{Bmatrix}x\,-\,1=&\Rightarrow=&x=&1 \\ \\ \\ 3x\,+\,1=&\Rightarrow=&x=$

 October 3rd, 2011, 03:41 AM #6 Newbie   Joined: Oct 2011 Posts: 10 Thanks: 0 Re: Solutions of the equation Thanks @soroban. Nice solution!!

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