Solve the equation below. List all the solutions. Include all complex solutions. 8x^3−27=0

Question

Help me out please! I have no idea how to solve it.

Answer 1

Once you have used your difference of cubes to factor

8x³-27=(2x-3)(4x²+6x+9)

you know that x=3/2 is a solution

Then you need to find the other solutions. If we only cared about real roots we would use the discriminant to see if it was worth solving the quadratic, but since we want complex roots as well we can just go ahead and commit to using the quadratic formula for the remaining 4x²+6x+9

a=4 b=6 c=9

(-6±√(36-4(4)(9)))/8

(-6±√(36-144))/8

(-6±√(-108))/8

√(-108)=i√(108)=i√(3*36)=6i√(3)

(-6±6i√(3))/8

(-3±3i√(3))/4

So our roots are one real root x=3/2 and two complex roots x=-3/4+3i√(3)/4 and x=-3/4-3i√(3)/4

Answer 2

The Difference of Cubes is

a³-b³=(a-b)(a²+ab+b²)

If we rewrite 8x³-27-0 like this: (2x)³-3³=0 and compare with (a²-b²)=(a-b)(a²+ab+b²) we find

a=2x, b=3 and the polynomial factors this way:

(2x-3)[(2x)²+(2x)(3)+3²]=

First solution: (2x-3)=0 giving x=3/2

Additional solutions:

(2x)²+(2x)(3)+3²=0

4x²+6x+9=0

The descriminant will tell us if there are any real roots to this:

If ax²+bx+c=0 and,

if b²-4ac>0 there are two real roots, and

if b²-4ac=0 there is one real root

if b²-4ac<0 there are no real roots

b²-4ac = 6²-4(4)(9) = 36-144 < 0 so there are no real roots

Solution: x=3/2