1. **Problem 474:** Simplify $$2(m - n)^2 (m + n) - 2(m + n)^2 (n - m) - 4m(m - n)(m + n)$$ 2. **Step 1:** Expand and rewrite terms. Note that $$n - m = -(m - n)$$, so $$- 2(m + n)^2 (n - m) = - 2(m + n)^2 (-(m - n)) = 2(m + n)^2 (m - n)$$ Rewrite expression: $$2(m - n)^2 (m + n) + 2(m + n)^2 (m - n) - 4m(m - n)(m + n)$$ 3. **Step 2:** Factor out common terms. Common factor is $$(m - n)(m + n)$$: $$= (m - n)(m + n) [2(m - n) + 2(m + n) - 4m]$$ 4. **Step 3:** Simplify inside the bracket: $$2(m - n) + 2(m + n) - 4m = 2m - 2n + 2m + 2n - 4m = (2m + 2m - 4m) + (-2n + 2n) = 0$$ 5. **Step 4:** Since the bracket is zero, the whole expression is zero. **Answer:** $$0$$ 2. **Problem 475:** Simplify $$ (a - 2b)^3 (a + 2b)^3 - (a^3 - 8b^3)(a^3 + 8b^3) - 12a^2 b^2 (2b + a)(-a + 2b) $$ 3. **Step 1:** Recognize identities. Note that $$(a - 2b)(a + 2b) = a^2 - 4b^2$$ So, $$(a - 2b)^3 (a + 2b)^3 = ((a - 2b)(a + 2b))^3 = (a^2 - 4b^2)^3$$ Also, $$(a^3 - 8b^3)(a^3 + 8b^3) = a^6 - (8b^3)^2 = a^6 - 64b^6$$ 4. **Step 2:** Rewrite expression: $$ (a^2 - 4b^2)^3 - (a^6 - 64b^6) - 12a^2 b^2 (2b + a)(-a + 2b) $$ 5. **Step 3:** Expand the last term: $$(2b + a)(-a + 2b) = 2b(-a) + 2b(2b) + a(-a) + a(2b) = -2ab + 4b^2 - a^2 + 2ab = 4b^2 - a^2$$ Simplify: $$-2ab + 2ab = 0$$ So, $$12a^2 b^2 (4b^2 - a^2) = 12a^2 b^2 imes 4b^2 - 12a^2 b^2 imes a^2 = 48 a^2 b^4 - 12 a^4 b^2$$ 6. **Step 4:** Expand $$(a^2 - 4b^2)^3$$ using binomial expansion: $$(a^2)^3 - 3(a^2)^2 (4b^2) + 3(a^2)(4b^2)^2 - (4b^2)^3 = a^6 - 3 a^4 (4b^2) + 3 a^2 (16 b^4) - 64 b^6$$ Simplify: $$a^6 - 12 a^4 b^2 + 48 a^2 b^4 - 64 b^6$$ 7. **Step 5:** Substitute back: $$ (a^2 - 4b^2)^3 - (a^6 - 64b^6) - 12a^2 b^2 (2b + a)(-a + 2b) = (a^6 - 12 a^4 b^2 + 48 a^2 b^4 - 64 b^6) - (a^6 - 64 b^6) - (48 a^2 b^4 - 12 a^4 b^2) $$ 8. **Step 6:** Simplify step by step: $$= a^6 - 12 a^4 b^2 + 48 a^2 b^4 - 64 b^6 - a^6 + 64 b^6 - 48 a^2 b^4 + 12 a^4 b^2$$ Group like terms: $$ (a^6 - a^6) + (-12 a^4 b^2 + 12 a^4 b^2) + (48 a^2 b^4 - 48 a^2 b^4) + (-64 b^6 + 64 b^6) = 0$$ 9. **Answer:** $$0$$ 3. **Problem 476:** Simplify $$ (x^2 + x + 1)(x^2 + x - 1) - (x^2 - 1)^2 - 3(x - 1)(x + 1) - 2^0 $$ 4. **Step 1:** Expand each term. First term: $$(x^2 + x + 1)(x^2 + x - 1) = x^2(x^2 + x - 1) + x(x^2 + x - 1) + 1(x^2 + x - 1)$$ $$= x^4 + x^3 - x^2 + x^3 + x^2 - x + x^2 + x - 1$$ Combine like terms: $$x^4 + 2x^3 + ( - x^2 + x^2 + x^2 ) + (-x + x) - 1 = x^4 + 2x^3 + x^2 - 1$$ 5. **Step 2:** Expand second term: $$(x^2 - 1)^2 = x^4 - 2x^2 + 1$$ 6. **Step 3:** Expand third term: $$(x - 1)(x + 1) = x^2 - 1$$ So, $$-3(x - 1)(x + 1) = -3(x^2 - 1) = -3x^2 + 3$$ 7. **Step 4:** Note that $$2^0 = 1$$ 8. **Step 5:** Substitute all back: $$ (x^4 + 2x^3 + x^2 - 1) - (x^4 - 2x^2 + 1) - 3x^2 + 3 - 1 $$ 9. **Step 6:** Simplify: $$x^4 + 2x^3 + x^2 - 1 - x^4 + 2x^2 - 1 - 3x^2 + 3 - 1$$ Group like terms: $$ (x^4 - x^4) + 2x^3 + (x^2 + 2x^2 - 3x^2) + (-1 - 1 + 3 - 1) = 0 + 2x^3 + 0 + 0 = 2x^3$$ 10. **Answer:** $$2x^3$$ 4. **Problem 477:** Simplify $$ (3c - 2z)^3 - (2c - 3z)^3 - (c + z)(19c^2 - 37cz + 19z^2) $$ 5. **Step 1:** Use the identity for difference of cubes: $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$ Let $$a = 3c - 2z$$ and $$b = 2c - 3z$$ Calculate $$a - b = (3c - 2z) - (2c - 3z) = c + z$$ Calculate $$a^2 + ab + b^2$$: - $$a^2 = (3c - 2z)^2 = 9c^2 - 12cz + 4z^2$$ - $$ab = (3c - 2z)(2c - 3z) = 6c^2 - 9cz - 4cz + 6z^2 = 6c^2 - 13cz + 6z^2$$ - $$b^2 = (2c - 3z)^2 = 4c^2 - 12cz + 9z^2$$ Sum: $$a^2 + ab + b^2 = (9c^2 - 12cz + 4z^2) + (6c^2 - 13cz + 6z^2) + (4c^2 - 12cz + 9z^2)$$ $$= (9 + 6 + 4) c^2 + (-12 - 13 - 12) cz + (4 + 6 + 9) z^2 = 19 c^2 - 37 cz + 19 z^2$$ 6. **Step 2:** Substitute back: $$ (3c - 2z)^3 - (2c - 3z)^3 = (a - b)(a^2 + ab + b^2) = (c + z)(19 c^2 - 37 cz + 19 z^2) $$ 7. **Step 3:** The expression is: $$ (3c - 2z)^3 - (2c - 3z)^3 - (c + z)(19 c^2 - 37 cz + 19 z^2) = (c + z)(19 c^2 - 37 cz + 19 z^2) - (c + z)(19 c^2 - 37 cz + 19 z^2) = 0 $$ 8. **Answer:** $$0$$