8x^6 -27=0?

Question

In my math class my teacher told me there was only one way to do this problem and that was through factoring and it was really long
but i said what if you took the cubic root of both sides you would get 8xsquared -3=0 wich eventually carried out to 1.2247 is this not a lidget way to do this problem? she would not listen to me she said the only way to do this problem is through factoring. am i right or is she right

The answer is actually + or - 1.2247

Answer 1

your teacher is right. your answer doesn't work because the cube root of 8x^6 - 27 is not 8x^2 - 3.

you need to factor according the rule for the difference of two cubes. The correct factoring is:
(2x^2 - 3)(4x^4 + 2x +9) = 0

Your solution in the real number system is:
2x^2 - 3 = 0
2x^2 = 3
x^2 = 3/2
x = sqrt(3/2) which should be simplified to:
x = (sqrt6)/2

If you set the other factor to zero, you can use the quadratic formula to find the complex solutions (which include imaginary numbers) to this equation.

Answer 2

You don't need to go through factoring if you only want to find real solutions. Here is one approach without factoring.

8x^6 =27
x^6 = 27/8 = (3/2)^3
x^2 = 3/2
x = +/-(3/2)^(1/2)

Answer 3

She is right, sorry :)
(8x^2 - 3) x (8x^2 -3) x (8x^2 - 3) does not = 8x^6 - 27.

Try it out. Neither does:

(2x^2 - 3)^3.

You have to factor, sorry.

Answer 4

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