f(x) = 8x³ + 27
To find the zeros equal the equation to 0
8x³ + 27 = 0
8x³ = -27
x³ = -27/8
x = [tex]\sqrt[3]{\frac{-27}{8}}[/tex]
x = [tex]\frac{-3}{2}[/tex]
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 = [tex]\sqrt[3]{f(z)}[/tex]
x = [tex]\sqrt[3]{f(z)}.\frac{-1-\sqrt{3}i}{2}[/tex]
x = [tex]\sqrt[3]{f(z)}.\frac{-1+\sqrt{3}i}{2}[/tex]
Making some subs.
x = [tex]\frac{-3}{2}[/tex]
x = [tex]\frac{-3}{2}.\frac{-1-\sqrt{3}i}{2}[/tex]
x = [tex]\frac{-3}{2}.\frac{-1+\sqrt{3}i}{2}[/tex]
So, after all we have:
x = [tex]\frac{-3}{2}[/tex]
x = [tex]\frac{3+3\sqrt{3}i}{4}[/tex]
x = [tex]\frac{3-3\sqrt{3}i}{4}[/tex]
