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#1

$Old$ November 19th, 2008, 05:20 AM

Alienis Back $Alienis Back is offline$

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$Default$ Need to facto this.

factoRRRR this...and I don't know where to start

$y^6-1$

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#2

$Old$ November 19th, 2008, 05:40 AM

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Originally Posted by Alienis Back $View Post$

factoRRRR this...and I don't know where to start

$y^6-1$

$= (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....$

And you should know how to factorise the sum and difference of two perfect cubes.

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#3

$Old$ November 19th, 2008, 07:08 AM

Alienis Back $Alienis Back is offline$

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Do you mean this??

$= (y^2)^3 - 1^3$

Where $a=y^2$ and $b=1$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

Then we have...

$(y^2)^3-1^3=(y^2-1) [(y^2)^2+y^2+1)]$

$(y^2)^3-1^3=(y^2-1) (y^4+y^2+1)$

...But I don't think this is the answer...mmm...I really don't

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#4

$Old$ November 19th, 2008, 03:42 PM

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Originally Posted by Alienis Back $View Post$

Do you mean this??

$= (y^2)^3 - 1^3$

Where $a=y^2$ and $b=1$

$a^3-b^3=(a-b)(a^2+ab+b^2)$

[snip]

How can you possibly think I mean that when I clearly wrote this:

Quote:

$= (y^3)^2 - 1^2 = (y^3 - 1)(y^3 + 1) \, ....$

And you should know how to factorise the sum and difference of two perfect cubes.

Now all you have to do is factorise each of $y^3 - 1 = y^3 - 1^3$ and $y^3 + 1 = y^3 + 1^3$ .

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The greater danger for most of us lies not in setting our aim too high and falling short; but in setting our aim too low and achieving our mark. (Michelangelo Buonarroti)

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