Real solutions

Let $y = 3x - 8$ , then this equation becomes $f(y) = y^3 - 264y - 1843 = 0$ with the same number of solutions.

If $f(y)$ is monotonous, it can only have one zero. If it is not monotonous, it will have a local maximum and a local minimum, and there may be one or three zeroes. The latter is true iff the maximum is positive and the minimum is negative.

The maximum and minimum of $f$ are found where $f'(y) = 3y^2 - 264 = 0$ , i.e. $y = \pm\sqrt{88}$ . The local maximum therefore has value

$f(-\sqrt{88}) = -(\sqrt{88})^3 + 264\sqrt{88} - 1543 = 176\sqrt{88} - 1843 < 176\cdot 10 - 1843 = -83 < 0.$

Because the local maximum is negative, the function will only cross zero after passing the local minimum. There is but $\boxed{\text{one}}$ zero.

Just use the formula and then analyse the results. There should be x 1=9 x 2=-1/2 + sqrt(3)/2 i x_3=-1/2 - sqrt(3)/2 i