2007-09-22 10:50:20 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x³-8=0
x³=8
x³=2*2*2
x=2
2007-09-22 11:00:55 · answer #1 · answered by Patty C 3 · 0⤊ 0⤋
x^3 - 8 = 0
x^3 - 2^3 = 0
Factorise this difference of two cubes :
(x - 2)(x^2 + 2x + 4) = 0
Set each term equal to zero :
x - 2 = 0 implies x = 2.
x^2 + 2x + 4 = 0 implies x = {-2 ± sqrt[2^2 - 4(1)(4)]} / (2*1)
using the quadratic formula.
Simplifying gives : x = -1 ± sqrt(3)*i
Solutions are : x = 2 or x = -1 - sqrt(3)*i or x = -1 + sqrt(3)*i
2007-09-22 17:59:21 · answer #2 · answered by falzoon 7 · 0⤊ 0⤋
Its of degree 3 so you know there's three such x's.... x=2 is obvious since 2^3=8... So divide x^3-8 by x-2 to get x^2 + 2x + 4 which has degree 2 so you can now just use the quadratic formula to get the last two possible x's...
x= -1 + sqrt(3)*i or x = -1 - sqrt(3)*i
2007-09-22 17:57:35 · answer #3 · answered by kuxuru 3 · 0⤊ 0⤋
x^3 - 8 is the difference of cubes.
(x - 2)(x^2 + 2x + 4)
x = 2 or x = 2i
2007-09-22 17:55:30 · answer #4 · answered by Anonymous · 0⤊ 0⤋
x^3 - 8 = 0
x^3 = 8
x = cube rt 8
x = 2
2007-09-22 17:59:46 · answer #5 · answered by lenpol7 7 · 0⤊ 0⤋