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

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b

answer: you need to organize this better it does not make sense.

step-by-step explanation: ?