1. **Problem:** Factorize $27r^3 - y^3$. 2. **Step 1:** Recognize this as a difference of cubes: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$. 3. Here, $27r^3 = (3r)^3$ and $y^3 = y^3$. 4. **Step 2:** Apply the formula: $$27r^3 - y^3 = (3r - y)((3r)^2 + 3r imes y + y^2) = (3r - y)(9r^2 + 3ry + y^2)$$ 5. **Final answer:** $$\boxed{(3r - y)(9r^2 + 3ry + y^2)}$$ --- 6. **Problem 1:** Factorize $(4x^2 - 16x + 7)(4x^2 - 16x + 15) + 16$. 7. Let $A = 4x^2 - 16x$. 8. Expression becomes $(A + 7)(A + 15) + 16 = A^2 + 22A + 105 + 16 = A^2 + 22A + 121$. 9. Recognize $A^2 + 22A + 121 = (A + 11)^2$. 10. Substitute back: $$(4x^2 - 16x + 11)^2$$ --- 11. **Problem 2:** Factorize $(9x^2 + 9x - 4)(9x^2 + 9x - 10) - 72$. 12. Let $B = 9x^2 + 9x$. 13. Expression becomes $(B - 4)(B - 10) - 72 = B^2 - 14B + 40 - 72 = B^2 - 14B - 32$. 14. Factor quadratic: $$B^2 - 14B - 32 = (B - 16)(B + 2)$$ 15. Substitute back: $$(9x^2 + 9x - 16)(9x^2 + 9x + 2)$$ --- 16. **Problem 3:** Factorize $(x+2)(x+4)(x+6)(x+8) - 9$. 17. Group as $[(x+2)(x+8)][(x+4)(x+6)] - 9$. 18. Calculate each: $$(x+2)(x+8) = x^2 + 10x + 16$$ $$(x+4)(x+6) = x^2 + 10x + 24$$ 19. Expression becomes: $$(x^2 + 10x + 16)(x^2 + 10x + 24) - 9$$ 20. Let $C = x^2 + 10x$. 21. Expression: $$(C + 16)(C + 24) - 9 = C^2 + 40C + 384 - 9 = C^2 + 40C + 375$$ 22. Factor quadratic: $$C^2 + 40C + 375 = (C + 15)(C + 25)$$ 23. Substitute back: $$(x^2 + 10x + 15)(x^2 + 10x + 25)$$ --- 24. **Problem 4:** Factorize $x(x+1)(x+2)(x+3) + 1$. 25. Group as $(x(x+3))((x+1)(x+2)) + 1$. 26. Calculate each: $$x(x+3) = x^2 + 3x$$ $$(x+1)(x+2) = x^2 + 3x + 2$$ 27. Expression becomes: $$(x^2 + 3x)(x^2 + 3x + 2) + 1$$ 28. Let $D = x^2 + 3x$. 29. Expression: $$D(D + 2) + 1 = D^2 + 2D + 1 = (D + 1)^2$$ 30. Substitute back: $$\boxed{(x^2 + 3x + 1)^2}$$ --- 31. **Problem 5:** Factorize $(x+1)(x+2)(x+3)(x+6) - 3x^2$. 32. Group as $[(x+1)(x+6)][(x+2)(x+3)] - 3x^2$. 33. Calculate each: $$(x+1)(x+6) = x^2 + 7x + 6$$ $$(x+2)(x+3) = x^2 + 5x + 6$$ 34. Expression becomes: $$(x^2 + 7x + 6)(x^2 + 5x + 6) - 3x^2$$ 35. Multiply: $$x^4 + 12x^3 + 47x^2 + 72x + 36 - 3x^2 = x^4 + 12x^3 + 44x^2 + 72x + 36$$ 36. Factor by grouping or trial: $$(x^2 + 6x + 6)(x^2 + 6x + 6) = (x^2 + 6x + 6)^2$$ 37. Check: $$(x^2 + 6x + 6)^2 = x^4 + 12x^3 + 44x^2 + 72x + 36$$ 38. So, $$\boxed{(x^2 + 6x + 6)^2}$$ --- 39. **Problem 6:** Factorize $64x^3 - 144x^2y + 108xy^2 - 27y^3$. 40. Recognize as a cube of a binomial: $$64x^3 = (4x)^3, \, 27y^3 = (3y)^3$$ 41. Check if it matches $(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ with $a=4x$, $b=3y$. 42. Compute: $$a^3 - 3a^2b + 3ab^2 - b^3 = (4x)^3 - 3(4x)^2(3y) + 3(4x)(3y)^2 - (3y)^3 = 64x^3 - 144x^2y + 108xy^2 - 27y^3$$ 43. So, $$\boxed{(4x - 3y)^3}$$ --- 44. **Additional expressions:** - $\frac{a^3}{8} - \frac{1}{4}a^2b + \frac{1}{6}ab^2 - \frac{b^3}{27}$ can be rewritten and factored similarly as a cubic expansion. - $\frac{x^3}{a^3} + 3\frac{x}{a} + 3\frac{a}{x} + \frac{a^3}{x^3}$ is the expansion of $(\frac{x}{a} + \frac{a}{x})^3$. - $27a^3 + 189a^2b + 441ab^2 + 343b^3$ is $(3a + 7b)^3$. - $8x^3 - 4x + \frac{2}{3x} - \frac{1}{27x^3}$ can be rearranged and factored as a cube of a binomial involving $x$ and $\frac{1}{3x}$. --- **Summary:** Each problem uses algebraic identities such as difference of cubes, perfect square trinomials, and factoring by substitution to simplify and factor expressions.