How to factor this equation?

0 ⤊

How to factor this equation?

27(K^3)-8

2006-10-18 14:55:52 · 3 answers · asked by FreakFreak 1 in Education & Reference ➔ Homework Help

3 answers

u cant factor that equation, it is prime

2006-10-18 14:57:52 · answer #1 · answered by Anonymous · 0⤊ 0⤋

It's a difference of cubes (3k)^3 - 2^3.

The formula for a difference of cubes is
A^3 - B^3 = (A - B)(A^2 +AB + B^2).

So you'd have (3k - 2)(9k^2 + 6k + 4)

2006-10-18 22:04:01 · answer #2 · answered by dmb 5 · 0⤊ 0⤋

27k^3- 8 =(3k)^3-(2)^3 = (3k- 2) {(3k)^2 +3k.2 +(2)^2] =(3k-2) (9k^2 +6k +4) ans [a^3-b^3=(a-b) (a^2 +ab =b^2) formula has been applied in the factorisation]

2006-10-19 05:42:48 · answer #3 · answered by alpha 7 · 0⤊ 0⤋