Help please!!!!!!!!?

Question

Solve 8x^3+1=0 using the sim of cubes formula

Answer 1

the sum on cubes formula factors this out to be (2x+1)(2x-1)(2x-1)=0

Then set each of the equations in the parenthesis equal to zero.

You come out with x= -1/2, 1/2

Answer 2

Sorry but yahoo answers won't be in your exams hun x

Answer 3

That made my brain hurt reading it.  Sorry I am of no service to ya dude.

Answer 4

Well that would translate into 8x^3+1^3. That would be

Formula is:

Sum of Cubes  -  An expression of the form a^3 + b^3. The sum of two cubes factors into (a + b)(a^2 - ab + b^2).

8+3( 8^2-8(1) +1^2)

That would then be

11(64-8+1)

11(57)

627

Answer 5

(2x + 1) (4x^2 - 2x + 1) = 0

2x + 1 = 0 and 4x^2 - 2x + 1 =0

x = -1/2 and ...... use quadratic formula to find the rest (too lazy to do the rest)

Answer 6

lol i hate math

Answer 7

This is the sum of cubes formula: x3 + y3 = (x+y)(x2-xy+y2)

Your equation can be rewritten as: (2x)^3 + 1^3 = 0

Now you can apply the sum of cubes formula and get:

(2x + 1) (4x^2 - 2x + 1) = 0

Which means 2x + 1 = 0 and thus x = -1/2

and also 4x^2 -2x + 1 = 0

We solve the quadratic equation using quadratic formula:

x1 = (2 + sqrt (4 - 16)) / 8 = (2 + sqrt(-12)) / 8

x2 = (2 - sqrt(-12)) / 8

So, the three roots are:

-1/2

(2 + sqrt(-12)) / 8

(2 - sqrt(-12)) / 8

Answer 8

i hate math

Answer 9

huh ? explain it please o.o