1. **State the problem:** Simplify the expression $$(3a+1)^2 - (3a+1)(2a-1)$$. 2. **Use the formula for expansion:** - Square of a binomial: $$(x+y)^2 = x^2 + 2xy + y^2$$ - Product of binomials: $$(x+y)(z+w) = xz + xw + yz + yw$$ 3. **Expand each term:** $$(3a+1)^2 = (3a)^2 + 2 \cdot 3a \cdot 1 + 1^2 = 9a^2 + 6a + 1$$ $$(3a+1)(2a-1) = 3a \cdot 2a + 3a \cdot (-1) + 1 \cdot 2a + 1 \cdot (-1) = 6a^2 - 3a + 2a - 1 = 6a^2 - a - 1$$ 4. **Subtract the second expression from the first:** $$9a^2 + 6a + 1 - (6a^2 - a - 1)$$ 5. **Distribute the minus sign:** $$9a^2 + 6a + 1 - 6a^2 + a + 1$$ 6. **Combine like terms:** $$9a^2 - 6a^2 + 6a + a + 1 + 1 = 3a^2 + 7a + 2$$ --- 1. **State the problem:** Simplify the expression $$2(a-2)^3 - (2-a)^2(5-a)$$. 2. **Note the relationship:** $$(2-a) = -(a-2)$$ 3. **Rewrite terms using this:** $$(2-a)^2 = [-(a-2)]^2 = (a-2)^2$$ $$(2-a)^3 = [-(a-2)]^3 = - (a-2)^3$$ 4. **Rewrite the expression:** $$2(a-2)^3 - (a-2)^2(5-a)$$ 5. **Expand the second factor:** $$5 - a = 5 - a$$ (leave as is for now) 6. **Factor out $(a-2)^2$ from both terms:** $$= (a-2)^2 \left[ 2(a-2) - (5 - a) \right]$$ 7. **Simplify inside the bracket:** $$2(a-2) - (5 - a) = 2a - 4 - 5 + a = 3a - 9$$ 8. **Final expression:** $$ (a-2)^2 (3a - 9)$$ 9. **Factor out 3 from the last term:** $$ (a-2)^2 \cdot 3(a - 3)$$ **Final simplified form:** $$3 (a-2)^2 (a-3)$$