Question

solve the equation; there will be three solutions.
27x^3-64=0

Answer 1

27x^3-64=0
(3x-4)(9x^2 + 12x + 16) =0
(3x-4) =0 or (9x^2 + 12x + 16) =0
Can you solve it from here?

Answer 2

do you just solve for x now?

Answer 3

Far to my knowledge, you cannot factorise (9x^2 + 12x + 16)
For that you need to use quadratics formula

Answer 4

I concur, I calculated 12x as 24x in my head...

Answer 5

x^3 = 64/27 -> x= 4/3...

Answer 6

Silly me.... I think too much :S

Answer 7

Still a problem though....

Answer 8

Best answer is from @estudier :)

Answer 9

ok i understand where the 4/3 comes from but not the 64/27

Answer 10

27x^3-64=0

Answer 11

Ho hum, 3 solutions.....

Answer 12

From (9x^2 + 12x + 16) =0, you can get 2 unreal solutions

Answer 13

64/27 comes from sending everything to the right :
27x^3-64 = 0
27x^3 = 64
x^3 = 64/27

Answer 14

OK, so you got 1 real from me and 2 complex from callisto, all done

Answer 15

(9x^2 + 12x + 16) =0
\[x=\frac{-12\pm \sqrt {12^2 -4(9)(16)}}{2(9)} = \frac{-12\pm \sqrt {-432}}{2(9)} =\frac{12(-1\pm \sqrt {-3})}{18}\]\[=\frac{2(-1\pm \sqrt {-3})}{3}\]