Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

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?

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,685

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

Offline

**im really bored****Member**- Registered: 2005-05-12
- Posts: 76

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

Offline

Pages: **1**