Solve: 27 - 8x^3
(3 -2x)(3+2x)^2
(3 -2x)(9+6x+4x^2)
is this right?
2006-09-13 17:40:46 · 9 answers · asked by rvn_villa 2 in Science & Mathematics ➔ Mathematics
Yes, that's correct.
(And, that's the difference of two cubes)
2006-09-13 17:42:40 · answer #1 · answered by Doug 2 · 0⤊ 1⤋
It is NOT a difference of two squares. It a is difference of two cubes! However, your answer is correct.
The formula for factoring is:
a^3 - b^3 = (a - b)(a^2 + a*b + b^2).
Now: 27 - 8x^3 = (3)^3 - (2*x)^3
= (3 - 2*x)*[(3)^2 + (3)*(2*x) + (2*x)^2]
= (3 - 2x)(9 + 6x + 4x^2)
2006-09-14 00:53:34 · answer #2 · answered by quidwai 4 · 0⤊ 0⤋
27 - 8x^3 = -8x^3 + 27 = -(8x^3 - 27) = -(2x - 3)(4x^2 + 6x + 9)
i prefer to put the variable in front. but yes you are right.
www.quickmath.com will also factor this for you.
2006-09-14 02:43:02 · answer #3 · answered by Sherman81 6 · 0⤊ 0⤋
Difference of two cubes is given by
a^3 - b^3 = (a - b) * (a^2 + ab + b^2)
In your case, a = 3 and b = 2x, so
... = (3 - 2x) (4x^2 + 6x + 9)
[which, by the way, is not equal to (3-2x)(3+2x)^2 !]
If you want to interpret this as the difference of two squares, you get
27 - 8x^3 = [3 sqrt 3 - 2x (sqrt 2x)] * [3 sqrt 3 + 2x (sqrt 2x)]
admittedly less elegant but correct.
2006-09-14 01:42:22 · answer #4 · answered by dutch_prof 4 · 0⤊ 0⤋
you been difference of 2 cubes
additionally
a^3-b^3 = (a-b)(a^2+ab+b^2) and not( a-b)(a+b)^2
based on this
it is (3-2x)(9+6x+4x^2)
2006-09-14 01:00:23 · answer #5 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
In the second part it would be 12x not 6x. That is the difference of two cubes, and that isn't the answer.
2006-09-14 00:50:50 · answer #6 · answered by Nelson_DeVon 7 · 0⤊ 0⤋
yes. but this is a difference of two cubes
2006-09-14 05:08:35 · answer #7 · answered by Anonymous · 0⤊ 0⤋
Yep, you're right. Good job :D
2006-09-14 00:43:30 · answer #8 · answered by b0b0link 2 · 0⤊ 1⤋
DAMN!!! The 'other' doug beat me to it âº
Doug
2006-09-14 00:45:36 · answer #9 · answered by doug_donaghue 7 · 0⤊ 1⤋