Find all solutions of the equation. Leave answers in trigono
\[
x^{3}+8=0
\]

Step 1 ：The given equation is \(x^{3}+8=0\).

Step 2 ：Rearranging the equation, we get \(x^{3} = -8\).

Step 3 ：The cube root of -8 is -2. However, this is only one solution. The other two solutions are complex numbers.

Step 4 ：To find these, we need to express -8 in polar form. The magnitude of -8 is 8, and the argument is π (since -8 lies on the negative real axis). Therefore, -8 can be expressed in polar form as \(8*(\cos(\pi) + i*\sin(\pi))\).

Step 5 ：According to De Moivre's theorem, the n-th roots of a complex number \(z = r*(\cos(\theta) + i*\sin(\theta))\) are given by \(z_k = r^{(1/n)} * (\cos((\theta + 2\pi k)/n) + i*\sin((\theta + 2\pi k)/n))\), for \(k = 0, 1, ..., n-1\). In this case, \(n = 3\), \(r = 8\), and \(\theta = \pi\).

Step 6 ：We can plug these values into the formula to find the three cube roots of -8.

Step 7 ：The roots of the equation are complex numbers. The real part of the roots is approximately 1 and -2, and the imaginary part is approximately ±1.732. These are the three solutions of the equation \(x^3 + 8 = 0\). The slight discrepancies from exact values are due to numerical precision errors in the calculations.

Step 8 ：Final Answer: The solutions of the equation are \(\boxed{1 + 1.732i, -2, 1 - 1.732i}\).