Consider the function f(x) = x3 -x - 8
This function is continuous
Let x = 2. Then f(2) = -1
Let x = 3. Then f(3) = 16
By the intermediate value theorem, on the interval
[2, 3], for f(2) < c < f(3) there is at least one x
such that f(x) = c.
Using the evaluated function values c would be in the
range -1 < c < 16.
Since 0 is in this interval, when c = 0, then f(x) = 0
has at least one solution, then x 3 = x + 8 has at
least one solution.