factor completely?

Question

8w^3 - 27


what would a solution be for a consistent independent line
Line 1: y = -x - 2
Line 2: y = - 1/2x - 5/2

Answer 1

8w^3 - 27

This is of the form x^3 - y^3
x^3 - y^3 = (x - y)(x^2 + xy + y^2)

8w^3 - 27 = (2w)^3 - 3^3
= (2w - 3)(4w^2 + 6w + 9)
= (2w - 3)[3(2w + 3) + 4w^2]

Answer 2

8w^3 - 27
= (2w - 3)(4w^2 + 6w + 9)


what would a solution be for a consistent independent line
Line 1: y = -x - 2
Line 2: y = - 1/2x - 5/2
-x-2=-1/2-5/2
-x/2 = -1/2
x=1
y= -1-2= -3

Answer 3

1) 8w^3 - 27
= (2w)^3 - (3)^3
=(2w-3)(4w^2 +6w +9)

but 8w^3 -27
= -21 (w^2 = 1)

2) y = -x - 2---------------(1)
y = -1/2x - 5/2-------------(2)

solving these two equations ,

=> -x -2 = -1/2x - 5/2

=> -1/2x = -1/2

=>x =1

putting the value of x in the equation (1)

y = -x -2
=> y = -1-2

=>y = -3

so the required solution is (1, -3).

Answer 4

1)8w^3-27
(2w-3)(4w^2+6w+9)
(2w-3)(2w+3)(2w+3)

2)Use the solve by graphing method:
-x-2=-1/2x-5/2

Multiply both sides by 2:
-2x-4=-x-5
-x-4=-5
-x=-1
x=1

y=-1-2
y=-3

The solution set is (1,-3)

I hope this helps!

Answer 5

8w^3 - 27 = (2w+3)(4w^2-9) = (2w+3)(2w+3)(2w-3)
--------------------------------

Line 1: y = -x - 2
Line 2: y = - 1/2x - 5/2
-x -2 = -x/2 -5/2
2x +4 = x+5
x=1