1. **Problem:** Factorise the expression $$x^2 - y^2 + yz - zx - 4x + 2y + z + 3$$. 2. **Formula and rules:** We use factoring techniques such as grouping, difference of squares, and rearranging terms to simplify and factor expressions. 3. **Step-by-step solution:** 1. Group terms to see patterns: $$x^2 - y^2 + yz - zx - 4x + 2y + z + 3 = (x^2 - y^2) + (yz - zx) - 4x + 2y + z + 3$$ 2. Recognize difference of squares: $$x^2 - y^2 = (x - y)(x + y)$$ 3. Factor the middle terms: $$yz - zx = z(y - x) = -z(x - y)$$ 4. Rewrite expression: $$(x - y)(x + y) - z(x - y) - 4x + 2y + z + 3$$ 5. Factor out $(x - y)$ from the first two terms: $$(x - y)(x + y - z) - 4x + 2y + z + 3$$ 6. Now focus on the remaining terms: $$-4x + 2y + z + 3$$ 7. Try to express the entire expression as a product of two binomials or factor further by grouping. Let's try grouping the last four terms: $$-4x + 2y + z + 3 = -4x + 2y + (z + 3)$$ 8. Notice that the first factor is $(x - y)$, so try to write the entire expression as: $$(x - y)(x + y - z) + A(x - y) + B$$ But since $A(x - y)$ is already factored, let's try to combine terms differently. 9. Alternatively, try grouping terms differently: $$x^2 - y^2 - zx + yz - 4x + 2y + z + 3$$ 10. Group as: $$(x^2 - zx - 4x) + (- y^2 + yz + 2y) + (z + 3)$$ 11. Factor each group: - From $x^2 - zx - 4x$, factor $x$: $$x(x - z - 4)$$ - From $- y^2 + yz + 2y$, factor $-y$: $$-y(y - z - 2)$$ - The last group is $z + 3$ 12. Notice $x - z - 4$ and $y - z - 2$ are similar but not equal, so no common factor here. 13. Try to rewrite the original expression as: $$(x - y)(x + y - z) - 4x + 2y + z + 3$$ 14. Let's test if the expression can be factored as: $$(x - y + a)(x + y - z + b)$$ Expanding: $$ (x - y)(x + y - z) + b(x - y) + a(x + y - z) + ab $$ From step 5, we have $(x - y)(x + y - z)$ already. So, the remaining terms $-4x + 2y + z + 3$ should equal: $$b(x - y) + a(x + y - z) + ab$$ 15. Equate coefficients: $$-4x + 2y + z + 3 = b(x - y) + a(x + y - z) + ab$$ Expand right side: $$b x - b y + a x + a y - a z + ab$$ Group terms: $$(a + b) x + (a - b) y - a z + ab$$ 16. Match coefficients: - Coefficient of $x$: $a + b = -4$ - Coefficient of $y$: $a - b = 2$ - Coefficient of $z$: $-a = 1 ightarrow a = -1$ - Constant term: $ab = 3$ 17. From $a = -1$, substitute into first two equations: - $-1 + b = -4 ightarrow b = -3$ - $-1 - b = 2 ightarrow -1 - (-3) = 2 ightarrow 2 = 2$ (checks out) 18. Check constant term: $$ab = (-1)(-3) = 3$$ correct. 19. Therefore, the factorization is: $$(x - y - 1)(x + y - z - 3)$$ --- **Final answer for (1):** $$\boxed{(x - y - 1)(x + y - z - 3)}$$ --- (2) **Problem:** Factorise the expression $$ac^2 - a^3 - a^2 b + ab^2 + b^3 - bc^2$$. **Step-by-step solution:** 1. Group terms to find common factors: $$(ac^2 - bc^2) + (-a^3 - a^2 b + ab^2 + b^3)$$ 2. Factor $c^2$ from the first group: $$c^2(a - b) + (-a^3 - a^2 b + ab^2 + b^3)$$ 3. Factor the second group by grouping: $$(-a^3 - a^2 b) + (ab^2 + b^3) = -a^2(a + b) + b^2(a + b)$$ 4. Factor $(a + b)$ from the second group: $$(-a^2 + b^2)(a + b)$$ 5. Recognize difference of squares: $$b^2 - a^2 = (b - a)(b + a)$$ 6. So, $$(-a^2 + b^2)(a + b) = (b - a)(b + a)(a + b) = (b - a)(a + b)^2$$ 7. Rewrite the entire expression: $$c^2(a - b) + (b - a)(a + b)^2$$ 8. Note that $b - a = -(a - b)$, so: $$c^2(a - b) - (a - b)(a + b)^2 = (a - b)(c^2 - (a + b)^2)$$ 9. Recognize difference of squares again: $$c^2 - (a + b)^2 = (c - (a + b))(c + (a + b)) = (c - a - b)(c + a + b)$$ 10. Therefore, the factorization is: $$(a - b)(c - a - b)(c + a + b)$$ --- **Final answer for (2):** $$\boxed{(a - b)(c - a - b)(c + a + b)}$$