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Old 11-19-2008, 05:20 AM
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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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Old 11-19-2008, 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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Old 11-19-2008, 07:08 AM
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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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Old 11-19-2008, 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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2. If you always do what you've always done, you'll always get what you've always got.

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4. Pressure makes diamonds.
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