Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

#1 2005-05-20 11:41:25

im really bored
Full Member

Offline

Some quick help please

factor completely:   128x^6  -  2y^6

I can expand it to:   2(8x^3 + y^3)(8x^3 - y^3)

I know it can be expanded more but how would that go?

#2 2005-05-20 15:53:29

MathsIsFun
Administrator

Offline

Re: Some quick help please

2(8x^3 + y^3)(8x^3 - y^3)

Hmm .... the two similar terms can be further factored like this

(2x+y)*(4x^2-2xy+y^2) ==> 8x^3 - 4x^2y + 2xy^2 + 4x^2y - 2xy^2 + y^3 ==> 8x^3 + y^3 
(2x-y)*(4x^2+2xy+y^2) ==> 8x^3 + 4x^2y + 2xy^2 - 4x^2y - 2xy^2 - y^3 ==> 8x^3 - y^3

So:

2(2x+y)*(4x^2-2xy+y^2)(2x-y)*(4x^2+2xy+y^2)


"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

#3 2005-05-21 08:46:39

im really bored
Full Member

Offline

Re: Some quick help please

Thanks, I was having some mental issues at the time...

Board footer

Powered by FluxBB