1. **State the problem:** Find the non-zero values of $a$ and $b$ such that $$(2x - a)^3 = 8x^3 - bx^2 + 2bx - a^3.$$ 2. **Recall the binomial expansion formula:** For any $u$ and $v$, $$(u - v)^3 = u^3 - 3u^2v + 3uv^2 - v^3.$$ 3. **Apply the formula to the left side:** Let $u = 2x$ and $v = a$, then $$ (2x - a)^3 = (2x)^3 - 3(2x)^2 a + 3(2x) a^2 - a^3 = 8x^3 - 12x^2 a + 6x a^2 - a^3.$$ 4. **Set the expanded left side equal to the right side:** $$8x^3 - 12x^2 a + 6x a^2 - a^3 = 8x^3 - b x^2 + 2b x - a^3.$$ 5. **Cancel $8x^3$ and $-a^3$ from both sides:** $$\cancel{8x^3} - 12x^2 a + 6x a^2 - \cancel{a^3} = \cancel{8x^3} - b x^2 + 2b x - \cancel{a^3}.$$ 6. **Simplify the equation:** $$-12x^2 a + 6x a^2 = -b x^2 + 2b x.$$ 7. **Group terms by powers of $x$:** $$(-12 a + b) x^2 + (6 a^2 - 2 b) x = 0.$$ 8. **For this to hold for all $x$, coefficients of $x^2$ and $x$ must be zero:** $$\begin{cases} -12 a + b = 0 \\ 6 a^2 - 2 b = 0 \end{cases}$$ 9. **Solve the system:** From the first equation: $$b = 12 a.$$ Substitute into the second: $$6 a^2 - 2 (12 a) = 0 \implies 6 a^2 - 24 a = 0.$$ 10. **Factor the second equation:** $$6 a (a - 4) = 0.$$ Since $a \neq 0$, $$a - 4 = 0 \implies a = 4.$$ 11. **Find $b$ using $b = 12 a$:** $$b = 12 \times 4 = 48.$$ **Final answer:** $$a = 4, \quad b = 48.$$