solve for z.\n\nz^{3}+64 = 0\n\nwrite your answers in simplified, rationalized form.

Explanation:

Step1: Rearrange the equation

$$z^{3}+64 = 0\Rightarrow z^{3}=-64$$

Step2: Use the formula (a^{3}+b^{3}=(a + b)(a^{2}-ab + b^{2}))

Here (z^{3}+64=z^{3}+4^{3}=(z + 4)(z^{2}-4z + 16)=0) From (z + 4=0), we get (z=-4) For (z^{2}-4z + 16=0), use the quadratic formula (z=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}) where (a = 1), (b=-4), (c = 16)

Step3: Calculate discriminant and roots

The discriminant (\Delta=b^{2}-4ac=(-4)^{2}-4\times1\times16=16 - 64=-48) (z=\frac{4\pm\sqrt{-48}}{2}=\frac{4\pm4\sqrt{3}i}{2}=2\pm2\sqrt{3}i)

Answer:

(z=-4), (z = 2 + 2\sqrt{3}i), (z=2-2\sqrt{3}i)