Question

Solve (y-6)^3+80=0 where y is a real number. Write your answer in simplified radical form.
(If there is more than one solution, separate them with commas)

Answer 1

\[(y-6)^3=-80\]
take the cubed root of both sides
\[y-6=\sqrt[3]{-80}\]
then add 6 to both sides
\[y=6+2i \sqrt[3]{10}, y=6-2i \sqrt[3]{10}\]]
i believe this is what your looking for