Question

Help me please... Find the real or imaginary solutions of this solution of each equation by factoring.
8x^3-27=0
Using: a^3-b^3=(a-b)(a^2+ab+b^2)

Answer 1

\[8x^3-27=(2x)^3-(3)^3\]

Answer 2

In your formula, replace each a with 2x and each b with 3

Answer 3

8x^3-27=0
(2x)^3 - (3)^(3) = 0
Using the formula,
(2x-3) ( 2x^2 + 6x +9) =0
(2x-3) = 0 or ( 2x^2 + 6x +9) =0
x=3/2 or 2[x^2 + 3x + (3/2)^2] - 2(3/2)^2 +9 =0
2(x + 3/2)^2 +27/2 = 0
2(x + 3/2)^2 = -27/2
(x + 3/2)^2 = -27/4
x+3/2 = (sqrt 27) i /2 or x+3/2 = - (sqrt 27) i /2
x= [(sqrt 27) -3] /2 or x = [-(sqrt 27)-3]/2

Answer 4

how did you find the second part?

Answer 5

the second part?

Answer 6

where is says x=3/2 from there on..

Answer 7

look at the (2x-3) = 0 first
(2x-3) = 0
2x = 3
x=3/2, got it?

Answer 8

ohhhhhh. yess

Answer 9

for ( 2x^2 + 6x +9) =0 part, you can either get the answer by completing square or using the quadratic formula

Answer 10

nah.. i guess i've made mistakes..

Answer 11

use the quad. formula,
\[(-6\pm \sqrt{6^{2}-4(2)(9)})/[2(2)] = (-6\pm \sqrt{-36}) /4\]\[ =(-6 \pm 6i) /4 \] \[ =(-3 \pm 3i) /2 \]

Answer 12

is there anything after that?