1. **State the problem:** Solve the equation $$8r^6 + 21sr^3 - 27 = 0$$ for $r$. 2. **Rewrite the equation:** Notice that $r^6 = (r^3)^2$. Let $x = r^3$. Then the equation becomes: $$8x^2 + 21sx - 27 = 0$$ 3. **Use the quadratic formula:** For a quadratic equation $ax^2 + bx + c = 0$, the solutions are: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=8$, $b=21s$, and $c=-27$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = (21s)^2 - 4 \times 8 \times (-27) = 441s^2 + 864$$ 5. **Find $x$:** $$x = \frac{-21s \pm \sqrt{441s^2 + 864}}{2 \times 8} = \frac{-21s \pm \sqrt{441s^2 + 864}}{16}$$ 6. **Recall substitution:** Since $x = r^3$, solve for $r$: $$r^3 = \frac{-21s \pm \sqrt{441s^2 + 864}}{16}$$ 7. **Take cube roots:** $$r = \sqrt[3]{\frac{-21s \pm \sqrt{441s^2 + 864}}{16}}$$ **Final answer:** $$r = \sqrt[3]{\frac{-21s + \sqrt{441s^2 + 864}}{16}} \quad \text{or} \quad r = \sqrt[3]{\frac{-21s - \sqrt{441s^2 + 864}}{16}}$$