1. **State the problem:** Solve the equation $$3\left(x+\frac{1}{x^2}\right)-7\left(1+\frac{1}{x}\right)=0$$ for $x$. 2. **Rewrite the equation:** $$3x + \frac{3}{x^2} - 7 - \frac{7}{x} = 0$$ 3. **Multiply through by $x^2$ to clear denominators:** $$3x^3 + 3 - 7x^2 - 7x = 0$$ 4. **Rearrange terms:** $$3x^3 - 7x^2 - 7x + 3 = 0$$ 5. **Try to factor the cubic polynomial:** Check for rational roots using factors of constant term 3: $\pm1, \pm3$. 6. **Test $x=1$:** $$3(1)^3 - 7(1)^2 - 7(1) + 3 = 3 - 7 - 7 + 3 = -8 \neq 0$$ 7. **Test $x=3$:** $$3(27) - 7(9) - 7(3) + 3 = 81 - 63 - 21 + 3 = 0$$ So, $x=3$ is a root. 8. **Divide polynomial by $(x-3)$:** Using polynomial division or synthetic division: $$3x^3 - 7x^2 - 7x + 3 = (x-3)(3x^2 + 2x - 1)$$ 9. **Solve quadratic $3x^2 + 2x - 1 = 0$ using quadratic formula:** $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-2 \pm \sqrt{4 + 12}}{6} = \frac{-2 \pm \sqrt{16}}{6} = \frac{-2 \pm 4}{6}$$ 10. **Find roots:** $$x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}$$ $$x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1$$ 11. **Final solutions:** $$x = 3, \quad x = \frac{1}{3}, \quad x = -1$$