64x^3-1=0
find the real or imaginary solutions by factoring

Question

internettroll11 · Answer

you have a 'difference of cubes' 
1=1^3 
64=4^3 

take (x^3-y^3) = (x-y)(x^2+xy+y^2) 

so 
(64x^3-1)=(4x-1)(16x^2+4x+1)

internettroll11 · Answer

the rest should be easy to do hopefully.

anonymous · Answer

Factoring: 64x3-1 To find the real root

Theory : A difference of two perfect cubes, a3 - b3 can be factored into
             (a-b) • (a2 +ab +b2)

Proof : (a-b)•(a2+ab+b2) =
            a3+a2b+ab2-ba2-b2a-b3 =
            a3+(a2b-ba2)+(ab2-b2a)-b3 =
            a3+0+0+b3 =
            a3+b3

Check : 64 is the cube of 4

Check : 1 is the cube of  1
Check : x3 is the cube of x1

Factorization is :
            (4x - 1) • (16x2 + 4x + 1) 
Solve for x, 4x-1= 0 

x=1/4 only real root

anonymous · Answer

Yet we still have to find imaginary roots, yes?

anonymous · Answer

@Nexxus do you know how to find those from the info. given? Or still need help :)