27x^3 – 8 factored
2007-03-25 17:25:29 · 7 answers · asked by Chris W 1 in Science & Mathematics ➔ Mathematics
difference of the cubic
a^3 - b^3 = (a-b)(a^2 + ab + b^2)
3^3x^3 - 2^3
(3x)^3 - 2^3
(3x - 2) (9x^2 + 6x + 4)
2007-03-25 17:29:35 · answer #1 · answered by 7 · 1⤊ 1⤋
27x^3 - 8 = (3x - 2)(9x^2 + 6x + 4)
2007-03-26 01:17:52 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
That is easy. This is the difference of a perfect cube. (3x-2)(9x^2+6x+4)
2007-03-26 00:32:28 · answer #3 · answered by Kristian 2 · 1⤊ 0⤋
(3x-2)(9x^2 + 6x + 4)
there's a formula for this. like there's the (a^2-b^2)=(a+b)(a-b)
this one's (a^3 - b^3) = (a-b)(a^2 + ab + b^2).
and (a^3 + b^3) = (a+b)(a^2 - ab + b^2)
Hope that helped
2007-03-26 00:32:57 · answer #4 · answered by Anonymous · 1⤊ 0⤋
a³ - b³ = (a - b).(a² + ab + b²)
a = 3x and b = 2
a³ - b³ = (3x - 2).(9x² + 6x + 4)
2007-03-26 05:25:18 · answer #5 · answered by Como 7 · 0⤊ 0⤋
all of them are right except for Juni Mcc
2007-03-26 00:35:07 · answer #6 · answered by Ha!! 2 · 0⤊ 0⤋
(3x-2)^3=0
x=2/3
I hope this helps. :)
2007-03-26 00:30:21 · answer #7 · answered by Juni Mccoy 3 · 0⤊ 2⤋