Chapter 8.4, Problem 41PS

Recall from the introduction to Section 8.2 that Jerome Cardans solutions to the equation $x^3=15x+4$ could be written as $x=\sqrt[3]{2+11i}+\sqrt[3]{2-11i}$

Let’s assume that the two cube roots are complex conjugates. If they are, then we can simplify our work by noticing that $x=\sqrt[3]{2+11i}+\sqrt[3]{2-11i}=a+bi+a-bi=2a$

which means that we simply double the real part of each cube root of $2+11i$ to find the solutions to $x^3=15x+4$. Now, to end our work with Cardan, find the three cube roots of $2+11i$. Then, noting the discussion above, use the three cube roots to solve the equation $x^3=15x+4$. Write your answers accurate to the nearest thousandth.