Question:

**Step 1:** Define the terms for the algebraic identity. Let $A = x-1$, $B = 2x-1$, and $C = -(3x-2)$. **Step 2:** Verify the condition for the algebraic identity. Calculate the sum $A+B+C$: $$A+B+C = (x-1) + (2x-1) + (-(3x-2))$$ $$A+B+C = x-1 + 2x-1 - 3x+2$$ $$A+B+C = (x+2x-3x) + (-1-1+2)$$ $$A+B+C = 0 + 0$$ $$A+B+C = 0$$ Since $A+B+C=0$, the algebraic identity $A^3+B^3+C^3 = 3ABC$ applies. **Step 3:** Apply the identity to the numerator of the given expression. The numerator of the given expression is $(x-1)^3 + (2x-1)^3 - (3x-2)^3$. This can be rewritten as $(x-1)^3 + (2x-1)^3 + (-(3x-2))^3$. Using the identity with $A=(x-1)$, $B=(2x-1)$, and $C=-(3x-2)$: $$(x-1)^3 + (2x-1)^3 + (-(3x-2))^3 = 3(x-1)(2x-1)(-(3x-2))$$ **Step 4:** Substitute the simplified numerator back into the original expression and simplify. The given expression is: $$\dfrac{(x-1)^3 + (2x-1)^3 - (3x-2)^3}{(x-1)(2x-1)(3x-2)}$$ Substitute the result from Step 3 into the numerator: $$ = \dfrac{3(x-1)(2x-1)(-(3x-2))}{(x-1)(2x-1)(3x-2)}$$ Assuming $(x-1)(2x-1)(3x-2) \neq 0$, the common terms $(x-1)$, $(2x-1)$, and $(3x-2)$ cancel out: $$ = 3 \cdot (-1)$$ $$ = -3$$