1. **State the problem:** Find the non-zero values of $a$ and $b$ such that $$(2x - a)^3 = 8x^3 - bx^2 + \frac{3}{2} b x - a^3.$$ 2. **Recall the binomial expansion formula:** $$(x - y)^3 = x^3 - 3x^2 y + 3x y^2 - y^3.$$ 3. **Apply the formula to $(2x - a)^3$:** $$(2x - a)^3 = (2x)^3 - 3(2x)^2 a + 3(2x) a^2 - a^3 = 8x^3 - 12 a x^2 + 6 a^2 x - a^3.$$ 4. **Match the given expression:** Given: $$8x^3 - b x^2 + \frac{3}{2} b x - a^3.$$ From expansion: $$8x^3 - 12 a x^2 + 6 a^2 x - a^3.$$ 5. **Equate coefficients of like powers of $x$: ** For $x^2$: $$-b = -12 a \implies b = 12 a.$$ For $x$: $$\frac{3}{2} b = 6 a^2.$$ 6. **Substitute $b = 12 a$ into the $x$ coefficient equation:** $$\frac{3}{2} (12 a) = 6 a^2$$ $$18 a = 6 a^2$$ Divide both sides by $6 a$ (since $a \neq 0$): $$\cancel{6 a} \times 3 = \cancel{6 a} \times a$$ $$3 = a.$$ 7. **Find $b$ using $a = 3$: ** $$b = 12 a = 12 \times 3 = 36.$$ **Final answer:** $$a = 3, \quad b = 36.$$