Factor the difference of the two cubes x^3-27?

Answer 1

*Use the difference of cubes formula which is >>>

(a - b)(a^2 + ab + b^2)

First: express the terms in lowest terms...

(x)(x)(x) - (3)(3)(3)

Sec: you have two variables...
a = x
b = 3

Third: replace the terms with their corresponding variables...

(x - 3)(x^2 + x(3) + 3^2)

(x - 3)(x^2 + 3x + 3*3)

(x - 3)(x^2 + 3x + 9)

Answer 2

bear in mind that the formula for the sum of two cubes is: a³ + b³ = (a + b)(a² - ab + b²) enable: a³ = sixty 4 a = sixty 4^(a million/3) a = 4 b³ = a³ b = a So the criteria are: (4 + a)(4² - 4a + a²) ==> (4 + a)(16 - 4a + a²) i'm hoping this helps!

Answer 3

(x-3)(x^2 + 3x + 9)
http://www.purplemath.com/modules/specfact2.htm
To check, do synthetic or polynomial division.
http://www.purplemath.com/modules/polydiv2.htm
or just multiply it out.
I see you're getting different answers here. This is why you can't just get the answers from YA, you have to understand it so that you can know which answer is right, and you can get it too.
Be sure and check the links I've given. They have good explanations of this and many other math topics. In fact you might want to bookmark the purplemath homepage. It's good, and it's free.

Answer 4

well, doing it longhand, I'm going to "guess" that one of the roots is (x-3).

(x-3) goes into x^3 - 27...
start with x^2

so we get x^3 - 27 - (x-3)x^2 as the remainder = 3x^2 - 27...
(x-3) into 3x^2 - 27... 3x times

and remainder of 3x^2 - 27 - (3x)(x-3) = 9x - 27
(x-3) into 9x-27.... 9 times

.. no remainder, it seems.

so (x-3)(x^2 + 3x + 9) are the two roots
this gives 3 as a root, as well as +/- 3/2i * (sqrt(3) - 1) (note the imaginariness)

Answer 5

x³ - 27 = (x - 3)(x² + 3x + 9)

Answer 6

(x-3)(x+3)(x+3)