1. **Stating the problem:** Solve the equation $$273x^8 - 17x^{12} = 256$$ for $x$. 2. **Rearrange the equation:** Move all terms to one side: $$273x^8 - 17x^{12} - 256 = 0$$ 3. **Rewrite the equation:** Factor out the common power of $x^8$ to simplify: $$x^8 (273 - 17x^4) = 256$$ 4. **Substitute:** Let $y = x^4$. Then the equation becomes: $$x^8 = (x^4)^2 = y^2$$ So, rewriting: $$y^2 (273 - 17y) = 256$$ 5. **Expand and simplify:** $$273y^2 - 17y^3 = 256$$ Rewrite as: $$-17y^3 + 273y^2 - 256 = 0$$ Or equivalently: $$17y^3 - 273y^2 + 256 = 0$$ 6. **Solve the cubic equation:** We want roots of $$17y^3 - 273y^2 + 256 = 0$$ Divide both sides by 17: $$y^3 - 16.0588235 y^2 + 15.0588235 = 0$$ 7. **Try to find rational roots via Rational Root Theorem:** The possible rational roots are factors of $15.0588235$ (approximate) over factors of 1. Try $y=1$: $$1 - 16.0588 + 15.0588 = 0$$ Approximately zero, so $y=1$ is a root. 8. **Factor out $(y - 1)$:** Divide cubic by $(y - 1)$ to get quadratic: $$17y^3 - 273y^2 + 256 = (y - 1)(17y^2 - 256y - 256)$$ Verify: division gives quadratic: $$17y^2 - 256y + 256$$ (There was a small miscalculation, let's do polynomial division step carefully.) Actual polynomial division: Divide $17y^3 - 273y^2 + 256$ by $y-1$: - First term: $17y^2$, multiply $(y-1)(17y^2) = 17y^3 - 17y^2$ - Subtract from original: $(17y^3 - 273y^2 + 256) - (17y^3 - 17y^2) = -256y^2 + 256$ - Next term: $-256y$, multiply $(y-1)(-256y) = -256y^2 + 256y$ - Subtract: $(-256y^2 + 256) - (-256y^2 + 256y) = -256y + 256$ - Next term: $-256$, multiply $(y-1)(-256) = -256y + 256$ - Subtract: $(-256y + 256) - (-256y + 256) = 0$ Therefore the factorization is: $$17y^3 - 273y^2 + 256 = (y - 1)(17y^2 - 256y + 256)$$ 9. **Solve quadratic equation:** $$17y^2 - 256y + 256 = 0$$ Use quadratic formula: $$y = \frac{256 \pm \sqrt{256^2 - 4 \times 17 \times 256}}{2 \times 17}$$ Calculate discriminant: $$256^2 = 65536$$ $$4 \times 17 \times 256 = 17408$$ So: $$\sqrt{65536 - 17408} = \sqrt{48128}$$ Calculate approximate value: $$\sqrt{48128} \approx 219.38$$ Thus, $$y = \frac{256 \pm 219.38}{34}$$ Two roots: $$y_1 = \frac{256 + 219.38}{34} = \frac{475.38}{34} \approx 13.98$$ $$y_2 = \frac{256 - 219.38}{34} = \frac{36.62}{34} \approx 1.08$$ 10. **Recall $y = x^4$:** So we have candidates: $$x^4 = 1$$ $$x^4 \approx 13.98$$ $$x^4 \approx 1.08$$ 11. **Solve for $x$:** - For $x^4 = 1$: $$x = \pm 1$$ - For $x^4 = 13.98$: $$x = \pm (13.98)^{1/4} \approx \pm 1.95$$ - For $x^4 = 1.08$: $$x = \pm (1.08)^{1/4} \approx \pm 1.02$$ 12. **Final solutions:** $$x = \pm 1, \pm 1.02, \pm 1.95$$