1. **State the problem:** Solve the equation $$2x + \frac{(x-1)^2 (128x) - (64x^2 + 2x - 1)}{(x-1)^4} = 0.$$\n\n2. **Rewrite the equation:** To combine terms, write $$2x$$ as $$\frac{2x (x-1)^4}{(x-1)^4}$$ to have a common denominator. The equation becomes:\n$$\frac{2x (x-1)^4 + (x-1)^2 (128x) - (64x^2 + 2x - 1)}{(x-1)^4} = 0.$$\n\n3. **Set numerator equal to zero:** Since denominator cannot be zero (except at $$x=1$$ which must be checked separately), solve:\n$$2x (x-1)^4 + (x-1)^2 (128x) - (64x^2 + 2x - 1) = 0.$$\n\n4. **Expand powers:**\n- $$(x-1)^2 = x^2 - 2x + 1$$\n- $$(x-1)^4 = ((x-1)^2)^2 = (x^2 - 2x + 1)^2 = x^4 - 4x^3 + 6x^2 - 4x + 1.$$\n\n5. **Substitute expansions:**\n$$2x (x^4 - 4x^3 + 6x^2 - 4x + 1) + (x^2 - 2x + 1)(128x) - (64x^2 + 2x - 1) = 0.$$\n\n6. **Distribute terms:**\n- $$2x (x^4 - 4x^3 + 6x^2 - 4x + 1) = 2x^5 - 8x^4 + 12x^3 - 8x^2 + 2x$$\n- $$(x^2 - 2x + 1)(128x) = 128x^3 - 256x^2 + 128x$$\n\n7. **Combine all terms:**\n$$2x^5 - 8x^4 + 12x^3 - 8x^2 + 2x + 128x^3 - 256x^2 + 128x - 64x^2 - 2x + 1 = 0.$$\n\n8. **Simplify by combining like terms:**\n- For $$x^5$$: $$2x^5$$\n- For $$x^4$$: $$-8x^4$$\n- For $$x^3$$: $$12x^3 + 128x^3 = 140x^3$$\n- For $$x^2$$: $$-8x^2 - 256x^2 - 64x^2 = -328x^2$$\n- For $$x$$: $$2x + 128x - 2x = 128x$$\n- Constant: $$+1$$\n\nSo the equation is:\n$$2x^5 - 8x^4 + 140x^3 - 328x^2 + 128x + 1 = 0.$$\n\n9. **Check for roots:** This is a quintic polynomial. Try rational root theorem candidates such as $$x=1$$:\n$$2(1)^5 - 8(1)^4 + 140(1)^3 - 328(1)^2 + 128(1) + 1 = 2 - 8 + 140 - 328 + 128 + 1 = -65 \neq 0.$$\nTry $$x=\frac{1}{2}$$ or other candidates or use numerical methods.\n\n10. **Check domain restriction:** The denominator is zero at $$x=1$$, so $$x \neq 1$$.\n\n**Final answer:** The solutions are the roots of $$2x^5 - 8x^4 + 140x^3 - 328x^2 + 128x + 1 = 0$$ excluding $$x=1$$.