2006-08-29 11:33:09 · 6 answers · asked by estyron2000 1 in Science & Mathematics ➔ Mathematics
x^6 - 8y^3 = (x^2 - 2y)(x^4 + 2x^2y + 4y^2)
2006-08-29 12:25:49 · answer #1 · answered by Sherman81 6 · 0⤊ 0⤋
A difference of cubes, a^3-b^3, factors as (a-b)(a^2+ab+b^2).
Write x^6-8y^3=(x^2)^3-(2y)^3 and use the difference of cubes formula with a=x^2 and b=2y:
x^6-8y^3=
(x^2)^3-(2y)^3)=
(x^2-2y)(x^4+2x^2y+4y^2)
Note that a sum of cubes also factors: a^3+b^3=(a+b)(a^2-ab+b^2). To make sure you understand, you should try to factor x^6+8y^3 on your own! :)
2006-08-30 08:12:05 · answer #2 · answered by Anonymous · 0⤊ 0⤋
It is actually in its simplest form. but if you set the terms = 0
X^6 -8y^3 = 0
Then you have
X^6 = 8y^3
Then you cube root
X^2 = 2y
2006-08-29 18:47:17 · answer #3 · answered by kryptoniam 1 · 0⤊ 1⤋
a^3 - b^3 = (a-b)(a^2 +ab +b^2)
in this problem a = x^2 and b= 2y
2006-08-29 23:09:24 · answer #4 · answered by qwert 5 · 0⤊ 0⤋
This function is not factorable. There are no common factors :)
2006-08-29 18:46:09 · answer #5 · answered by kaceelynn1987 2 · 0⤊ 1⤋
(X^2-8Y)^3
I think
2006-08-29 18:46:01 · answer #6 · answered by nedoglover 4 · 0⤊ 1⤋