1. State the problem: Solve the equation $12^x - 2 = 3^{3x} \cdot 2^{6x}$ for $x$.\n\n2. Express bases with prime factorization: $12 = 2^2 \cdot 3$, so $12^x = (2^2 \cdot 3)^x = 2^{2x} \cdot 3^x$. Rewrite the right side: $3^{3x} \cdot 2^{6x}$ stays as it is.\n\n3. Substitute: The equation becomes $$2^{2x} \cdot 3^x - 2 = 3^{3x} \cdot 2^{6x}.$$\n\n4. Express the right-hand side similarly: $3^{3x} = (3^x)^3$ and $2^{6x} = (2^{2x})^3$. So the right side is $(3^x)^3 \cdot (2^{2x})^3 = (3^x \cdot 2^{2x})^3$.\n\n5. Introduce substitution variables: Let $a = 3^x$ and $b = 2^{2x}$. Then the equation becomes: $$a b - 2 = (a b)^3.$$\n\n6. Rewrite as $$ (a b)^3 - a b + 2 = 0.$$ Let $y = a b$. The cubic equation is $$y^3 - y + 2 = 0.$$\n\n7. Solve the cubic equation $y^3 - y + 2 = 0$. Test possible rational roots $\pm1, \pm2$.\n\n8. Test $y = -1$: $$(-1)^3 - (-1) + 2 = -1 + 1 + 2 = 2 \neq 0.$$ Test $y = -2$: $$(-2)^3 - (-2) + 2 = -8 + 2 + 2 = -4 \neq 0.$$ Test $y = 1$: $$1 - 1 + 2 = 2 \neq 0.$$ Test $y = 2$: $$8 - 2 + 2 = 8 \neq 0.$$\n\n9. No rational roots found. Use the cubic formula or approximate solutions. The discriminant is $$\Delta = -4(-1)^3 - 27(2)^2 = -4(-1) - 27 \cdot 4 = 4 - 108 = -104 < 0,$$ so there is one real root.\n\n10. Use numerical methods (e.g., Newton's method) or approximate root: the real root is about $y \approx -1.7693$.\n\n11. Recall $y = ab = 3^x \cdot 2^{2x} = 3^x \cdot (2^2)^x = 3^x \cdot 4^x = (3 \cdot 4)^x = 12^x.$\n\n12. So $12^x = y \approx -1.7693$, but $12^x$ is always positive for real $x$, so no real solution exists.\n\n13. Conclusion: There is no real solution to the equation $12^x - 2 = 3^{3x} \cdot 2^{6x}$ because the equivalent cubic equation has only one real root which is negative and cannot equal $12^x$.