f(x) = 8x³ + 27
To find the zeros equal the equation to 0
8x³ + 27 = 0
8x³ = -27
x³ = -27/8
x =
![\sqrt[3]{(-27)/(8)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/bndg7fxhws4tskw9beqic8td9nbp59sck3.png)
x =

This is the real root, but there are two other irrational roots.
By rule we know:
For f(z) = x³ the three solutions are:
x =
![\sqrt[3]{f(z)}](https://img.qammunity.org/2019/formulas/mathematics/high-school/t4wnyn4ip125rhlr5m940i9e4bee190vr0.png)
x =
![\sqrt[3]{f(z)}.(-1-√(3)i)/(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/56ixxput7w279pb0q0rl87o0jaehc57vo3.png)
x =
![\sqrt[3]{f(z)}.(-1+√(3)i)/(2)](https://img.qammunity.org/2019/formulas/mathematics/high-school/ulmmk5kefw0b8y8c9pks06p4u8e3gactjx.png)
Making some subs.
x =

x =

x =

So, after all we have:
x =

x =

x =
