Pls help! how would the expression x^3-8 be rewritten using difference of cubes?

A- (x+2)(x^2+2x-4)
B- (x-2)(x^2-2x+4)
C-(x+2)(x^2-2x+4)
D-(x-2)(x^2+2x+4)

D

Step-by-step explanation:

Difference of cubes is as follows

x^3-y^3=(x-y)(x^2+y^2+xy)

*notice how the sign in the origonal equation is the same for the first part but then flipped for the second

8 can be rewritten as 2^3 so we have

x^3-2^3 which means that we can use the template from before

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

simplify the 2^2 into 4 and put the variables in front to get

(x-2)(x^2+2x+4)

This is the same as D

part a

x     y

0     8

1     7.5

2     7

3     6.5

4     6

5     5.5  

for every hour i deducted .5 inches from the length of the candle

part b

it is a function because for every x value there is only one y value

part   c

for every x value there is only one y value

x       y

0       8

1       7.6

2       7.2

3       6.8

step-by-step explanation:

linear pair postulates