1. **State the problem:** Solve the equation $$x^6 + 7x^3 - 8 = 0$$ for $x$. 2. **Use substitution:** Let $y = x^3$. Then the equation becomes $$y^2 + 7y - 8 = 0$$ because $x^6 = (x^3)^2 = y^2$. 3. **Solve the quadratic equation:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=7$, and $c=-8$. 4. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 7^2 - 4 \times 1 \times (-8) = 49 + 32 = 81$$. 5. **Find the roots for $y$:** $$y = \frac{-7 \pm \sqrt{81}}{2} = \frac{-7 \pm 9}{2}$$ 6. **Evaluate each root:** - For $+$ sign: $$y = \frac{-7 + 9}{2} = \frac{2}{2} = 1$$ - For $-$ sign: $$y = \frac{-7 - 9}{2} = \frac{-16}{2} = -8$$ 7. **Back-substitute $y = x^3$:** - When $y=1$, $$x^3 = 1 \implies x = \sqrt[3]{1} = 1$$ - When $y=-8$, $$x^3 = -8 \implies x = \sqrt[3]{-8} = -2$$ 8. **Final solutions:** $$x = 1 \text{ or } x = -2$$. These are the real roots of the original equation.