1. **State the problem:** Given the system of equations $$x^3 = 3 - y^3$$ and $$x^6 = 27 - y^6,$$ find the value of $x$. 2. **Recall the formulas and rules:** We know that $$x^6 = (x^3)^2$$ and $$y^6 = (y^3)^2.$$ This allows us to rewrite the second equation in terms of $x^3$ and $y^3$. 3. **Rewrite the second equation:** Substitute $$x^6 = (x^3)^2$$ and $$y^6 = (y^3)^2$$ into the second equation: $$ (x^3)^2 = 27 - (y^3)^2 $$ 4. **Use the first equation:** From the first equation, $$x^3 = 3 - y^3$$. Let $$a = x^3$$ and $$b = y^3$$, so $$a = 3 - b$$. 5. **Rewrite the second equation in terms of $a$ and $b$:** $$ a^2 = 27 - b^2 $$ 6. **Substitute $a = 3 - b$ into the above:** $$ (3 - b)^2 = 27 - b^2 $$ 7. **Expand and simplify:** $$ 9 - 6b + b^2 = 27 - b^2 $$ 8. **Bring all terms to one side:** $$ 9 - 6b + b^2 + b^2 - 27 = 0 $$ $$ 2b^2 - 6b - 18 = 0 $$ 9. **Divide entire equation by 2:** $$ \cancel{2}b^2 - 3b - 9 = 0 $$ $$ \cancel{2} \cancel{b^2} - 3b - 9 = 0 $$ 10. **Solve the quadratic equation:** $$ 2b^2 - 6b - 18 = 0 $$ Use quadratic formula: $$ b = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 2 \times (-18)}}{2 \times 2} = \frac{6 \pm \sqrt{36 + 144}}{4} = \frac{6 \pm \sqrt{180}}{4} $$ 11. **Simplify the square root:** $$ \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} $$ 12. **Calculate $b$ values:** $$ b = \frac{6 \pm 6\sqrt{5}}{4} = \frac{6}{4} \pm \frac{6\sqrt{5}}{4} = \frac{3}{2} \pm \frac{3\sqrt{5}}{2} $$ 13. **Find corresponding $a$ values:** Recall $$a = 3 - b$$ - For $$b = \frac{3}{2} + \frac{3\sqrt{5}}{2}$$: $$ a = 3 - \left(\frac{3}{2} + \frac{3\sqrt{5}}{2}\right) = \frac{3}{2} - \frac{3\sqrt{5}}{2} $$ - For $$b = \frac{3}{2} - \frac{3\sqrt{5}}{2}$$: $$ a = 3 - \left(\frac{3}{2} - \frac{3\sqrt{5}}{2}\right) = \frac{3}{2} + \frac{3\sqrt{5}}{2} $$ 14. **Recall $a = x^3$, so:** $$ x^3 = \frac{3}{2} \pm \frac{3\sqrt{5}}{2} $$ 15. **Final answer:** $$ x = \sqrt[3]{\frac{3}{2} + \frac{3\sqrt{5}}{2}} \quad \text{or} \quad x = \sqrt[3]{\frac{3}{2} - \frac{3\sqrt{5}}{2}} $$