1. **State the problem:** Factorise the expression $$(y+z-x)^3 + (z+x-y)^3 + (x+y-z)^3 = -24xyz$$ given that $$x + y + z = 0$$. 2. **Recall the identity:** There is a well-known factorization identity for cubes when the sum of variables is zero: $$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$$ If $$a + b + c = 0$$, then $$a^3 + b^3 + c^3 = 3abc$$. 3. **Apply the identity:** Let $$a = y + z - x$$ $$b = z + x - y$$ $$c = x + y - z$$ Notice that $$a + b + c = (y+z-x) + (z+x-y) + (x+y-z) = 2(x + y + z) = 0$$ (since $$x + y + z = 0$$). 4. **Use the identity:** Since $$a + b + c = 0$$, $$a^3 + b^3 + c^3 = 3abc$$. 5. **Rewrite the original expression:** $$(y+z-x)^3 + (z+x-y)^3 + (x+y-z)^3 = 3(y+z-x)(z+x-y)(x+y-z)$$ 6. **Given the original equation:** $$(y+z-x)^3 + (z+x-y)^3 + (x+y-z)^3 = -24xyz$$ From step 5, this equals $$3(y+z-x)(z+x-y)(x+y-z) = -24xyz$$ 7. **Divide both sides by 3:** $$(y+z-x)(z+x-y)(x+y-z) = -8xyz$$ **Final factorization:** $$ (y+z-x)^3 + (z+x-y)^3 + (x+y-z)^3 = 3(y+z-x)(z+x-y)(x+y-z) = -24xyz $$ and $$ (y+z-x)(z+x-y)(x+y-z) = -8xyz $$