Total no of real solution

I'm going to take a calculus approach here. Let $f(x) = x^3 - 8x^2 - 8x - 9$ , and $f'(x) = 3x^2 - 16x - 8 = 0 \Rightarrow x = \frac{16 \pm \sqrt{16^2 - 4(3)(-8)}}{6} = \frac{8 \pm 2\sqrt{22}}{3}.$ be the local extrema. We now find that $f(\frac{8-2\sqrt{22}}{3})$ and $f(\frac{8+2\sqrt{22}}{3})$ are both negative valued. The function will only cross the x-axis ( $y = 0$ ) in the interval $(\frac{8 + 2\sqrt{22}}{3}, +\infty)$ , which implies $f(x) = 0$ only has one real root.