1. **State the problem:** Solve the inequality $$(3x + 1)(1 - 3x) + 2(1 - 3x) \geq (x - 1)^3 - x^2(6 + x).$$ 2. **Expand and simplify each side:** - Left side: $$ (3x + 1)(1 - 3x) + 2(1 - 3x) = (3x)(1) + (3x)(-3x) + 1(1) + 1(-3x) + 2 - 6x = 3x - 9x^2 + 1 - 3x + 2 - 6x = -9x^2 - 6x + 3.$$ - Right side: $$ (x - 1)^3 - x^2(6 + x) = (x - 1)^3 - 6x^2 - x^3.$$ Expand $(x - 1)^3$ using binomial expansion: $$ (x - 1)^3 = x^3 - 3x^2 + 3x - 1.$$ So right side becomes: $$ x^3 - 3x^2 + 3x - 1 - 6x^2 - x^3 = -9x^2 + 3x - 1.$$ 3. **Rewrite the inequality:** $$ -9x^2 - 6x + 3 \geq -9x^2 + 3x - 1.$$ 4. **Bring all terms to one side:** $$ -9x^2 - 6x + 3 - (-9x^2 + 3x - 1) \geq 0,$$ which simplifies to $$ -9x^2 - 6x + 3 + 9x^2 - 3x + 1 \geq 0,$$ $$ (-9x^2 + 9x^2) + (-6x - 3x) + (3 + 1) \geq 0,$$ $$ -9x + 4 \geq 0.$$ 5. **Solve the linear inequality:** $$ -9x + 4 \geq 0 \implies -9x \geq -4.$$ Divide both sides by $-9$ and reverse inequality sign: $$ \cancel{-9}x \leq \cancel{-4} \div -9,$$ $$ x \leq \frac{4}{9}.$$ **Final answer:** $$ \boxed{x \leq \frac{4}{9}}.$$