does anybody know the answer to this x3 (cubed) + 27?

Answer 1

x^3+27
=(x)^3+(3)^3
=(x+3) {(x^2-x.3+(3)^2}
=(x+3)(x^2-3x+9)

Answer 2

How to factor a sum of cubes:

1) Take the cube root of each term and place in the first set of brackets. In this case, we're going to have
(x + 3)

2) From these values, you're going to construct values in a second set of brackets, which goes as follows:
a) Square the first
b) Negative product
c) Square the last.

"Square the first"

For (x - 3), you want to square the first value of the brackets. Therefore, you square x, which is x^2.

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

"Negative product"

For the terms x and -3, take their product by multiplying them together (-3x), and then take their *negative* by multiplying by (-1). So you should get 3x.

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

"Square the last"

This means to square the second term in the first set of brackets. As you can see, (-3)^2 = 9, so we have

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

You have now successfully factored a sum of cubes. Difference of cubes works the exact same way.

Answer 3

do you mean x^3+27=0?
if so the answer is -3

Answer 4

Your statement doesn't have an assertion (like an "="), so it has no answer.

if x^3 +27 = 0, then x=-3

Answer 5

Factoring? This is the sum of two cubes:
(x+3)(x^2-3x+9)

Answer 6

x^3+27=0
x^3=-27
x=3i
("i" is an imaginary # used as a substitute for negative 1

Answer 7

x3^3+27
27x+27
27x=-27
-1