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October 2nd, 2011, 08:03 AM   #1
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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..
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October 2nd, 2011, 08:16 AM   #2
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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
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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 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?
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October 2nd, 2011, 09:31 AM   #4
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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.
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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:

[color=beige]. . [/color]

Note that all the terms can be factored:

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[color=beige]. . [/color]

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





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




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

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

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




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October 3rd, 2011, 03:41 AM   #6
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Re: Solutions of the equation

Thanks @soroban. Nice solution!!
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