Question:

Step 1: Recognize the algebraic identity for sum of cubes. We need to evaluate the given expression. Notice that the numerator has the form $a^3 + b^3 + c^3$, which suggests using the algebraic identity: when $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$. Step 2: Define substitutions to apply the identity. Let us define: $$a = x-1$$ $$b = 2x-1$$ $$c = -(3x-2)$$ Step 3: Verify that $a + b + c = 0$. We compute: $$a + b + c = (x-1) + (2x-1) + (-(3x-2))$$ $$= x - 1 + 2x - 1 - 3x + 2$$ $$= (x + 2x - 3x) + (-1 - 1 + 2)$$ $$= 0$$ Since the sum equals zero, the identity $a^3 + b^3 + c^3 = 3abc$ applies. Step 4: Express the numerator using the identity. The numerator can be rewritten as: $$(x-1)^3 + (2x-1)^3 - (3x-2)^3 = a^3 + b^3 + c^3$$ Using the identity $a^3 + b^3 + c^3 = 3abc$: $$a^3 + b^3 + c^3 = 3(x-1)(2x-1)(-(3x-2))$$ $$= -3(x-1)(2x-1)(3x-2)$$ Step 5: Simplify the original expression. Now we can evaluate the complete expression: $$\frac{(x-1)^3 + (2x-1)^3 - (3x-2)^3}{(x-1)(2x-1)(3x-2)} = \frac{-3(x-1)(2x-1)(3x-2)}{(x-1)(2x-1)(3x-2)}$$ Canceling the common factor $(x-1)(2x-1)(3x-2)$ from numerator and denominator: $$= -3$$ **Final Answer:** The value of the expression is $\boxed{-3}$, which corresponds to **Option 1**.